Formula for the Sum and Difference of Two Angles in Trigonometry

This time we will discuss the formula for the sum and difference of two angles in trigonometry along with evidence and example questions

Trigonometry is Mathematics material that is always available at every grade level in high school, whether in class X, XI or XII. In class X there are ratios, functions and trigonometric identities. In Class XI there are Trigonometric Equations and Formulas for the Sum and Difference of two angles. Meanwhile, in class XII, you will encounter trigonometry when calculating indefinite integrals and definite integrals from trigonometric functions.

This time, we will learn together about Formulas for Sum and Difference of Angles in Trigonometry. This is advanced trigonometry material. This material is used to calculate trigonometric angle values ​​that are not special. For example $\sin 75^\circ$, we can write it as the sum of the special angles $\sin 75^\circ=\sin (45^\circ+30^\circ)$. For example, again $\cos 15^\circ$, we can write it by subtracting or differenting the special angle $\cos 15^\circ=\cos (45^\circ-30^\circ)$.

To make it easier to understand this material, of course you must first understand the trigonometry lesson in class Well, don’t forget this material, because this material is one of the prerequisites for understanding this module. Come on, don’t be afraid and don’t know how to do trigonometry, let’s learn gradually and be sure we can do it…

1. Formulas for $\cos(\alpha+\beta)$ and $\cos(\alpha-\beta)$

Pembuktian formula $\cos(\alpha+\beta)=\cos\alpha \cos\beta -\sin\alpha \sin\beta$

Let’s prove both formulas. There are 2 ways to prove it

Method 1: Using the unit circle, find the distance between two points

Consider the unit circle and 3 circles whose angles are $\alpha$, $\beta$, and $-\beta$.

The figure above is the unit circle with center O and radius $r$. Consider two triangles namely $\vartriangle POR$ and $\vartriangle SOQ$ with $\angle POR$ and $\angle SOQ$ such that $PR=SQ$.

By proving $PR=SQ$, we obtain $\cos(\alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ From this image, it is obtained
$OP=OQ=OR=OS = r$ The polar coordinates are: point $P(r, 0)$,
point $Q(r \cos \alpha, r \sin \alpha )$,
titik $R(r \cos(\alpha + \beta ), r \sin(\alpha + \beta ))$ , and point $S(r \cos \beta , -r \sin \beta )$.
Remember

1. Concept of distance between two points A($x_1,y_1$) and B($x_2,y_2$): $$\begin{align*} AB &= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 } \\ AB^2 &= (x_2-x_1)^2 + (y_2-y_1)^2 \end{align*}$$
2. Identitas trigonometri: $\sin ^2 A + \cos ^2 A = 1$

Pembuktian formula $\sin(\alpha+\beta)=\sin\alpha \cos\beta -\cos\alpha \sin\beta$
Finding PR Distance:
$P(r, 0)$ and $R (r \cos(\alpha + \beta ) , r \sin(\alpha + \beta ))$ with the concept of distance between two points above $$\begin{align*}PR^2 &= [r \cos(\alpha + \beta ) - r]^2 + [r \sin(\alpha + \beta ) - 0 ]^2 \\ &= [r \cos(\alpha + \beta ) - r]^2 + [r \sin(\alpha + \beta ) - 0 ]^2 \\ &= r^2 \cos ^2 (\alpha + \beta ) - 2r^2 \cos(\alpha + \beta ) + r^2 + r^2 \sin ^2 (\alpha + \beta ) \\ &= r^2 [\cos ^2 (\alpha + \beta ) + \sin ^2 (\alpha + \beta ) ]- 2r^2 \cos(\alpha + \beta ) + r^2 \\ &= r^2 [1 ]- 2r^2 \cos(\alpha + \beta ) + r^2\\ PR^2 &= 2r^2 - 2r^2 \cos(\alpha + \beta ) \end{align*}$$ **Finding QS Distance :**
$Q(r \cos \beta , -r \sin \beta )$ and $S(r \cos \alpha, r \sin \alpha )$ with the concept of distance between two points above $$\begin{align*}QS^2 &= [r \cos \alpha - r \cos \beta]^2 + [r \sin \alpha - ( -r \sin \beta ) ]^2 \\ &= [r \cos \alpha - r \cos \beta]^2 + [r \sin \alpha + r \sin \beta ]^2 \\ &= (r^2 \cos ^2 \alpha - 2r^2 \cos \alpha \cos \beta + r^2 \cos ^2 \beta ) + ( r^2 \sin ^2 \alpha + 2r^2 \sin \alpha \sin \beta + r^2 \sin ^2 \beta ) \\ &= r^2 (\cos ^2 \alpha + \sin ^2 \alpha ) + r^2 ( \cos ^2 \beta + \sin ^2 \beta) -2r^2 ( \cos \alpha \cos \beta - \sin \alpha \sin \beta ) \\ &= r^2 (1 ) + r^2 ( 1 ) -2r^2 ( \cos \alpha \cos \beta - \sin \alpha \sin \beta ) \\ QS^2 &= 2r^2 -2r^2 ( \cos \alpha \cos \beta - \sin \alpha \sin \beta ) \end{align*}$$ **The length of PR is the same as the length of QS** $$\begin{align*}PR &= QS \\PR^2 &= QS^2 \\2r^2 - 2r^2 \cos(\alpha + \beta ) &= 2r^2 -2r^2 ( \cos \alpha \cos \beta - \sin \alpha \sin \beta ) \\ \cos(\alpha + \beta ) &= \cos \alpha \cos \beta - \sin \alpha \sin \beta \end{align*}$$ So it is proven: $\cos(\alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

Pembuktian formula $\cos(\alpha - \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$

The formula $\cos(\alpha-\beta)$ can be obtained with the negative angle relation property, namely $\cos(\alpha+(-\beta))$
Negative angle concept: $\sin (-A) = - \sin A$ and $\cos ( -A) = \cos A$
$$\begin{align*} \cos(\alpha - \beta ) &= \cos(\alpha + (- \beta) ) \\ &= \cos \alpha \cos (-\beta) - \ sin \alpha \sin (- \beta ) \\ &= \cos \alpha \cos \beta - \sin \alpha . (- \sin \beta) \\ &= \cos \alpha \cos \beta + \sin \alpha \sin \beta \end{align*}$$ So it is proven: $\cos(\alpha - \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$

Method 2: Using the Cosine Rule and Distance of Two Points

Look at the following picture,
In the picture, for example, a unit circle with radius $r= 1$ and center O. Pay attention to $\triangle SOQ$, the polar coordinates are point S and point Q, namely $S(\cos \beta , -\sin \beta)$ dan $Q(\cos \alpha , \sin \alpha )$ serta OS = OQ = 1.
Remember
Identitas trigonometri: $\sin ^2 A + \cos ^2 A = 1$
Next, find the distance between points S and Q using the formula:
$$\begin{align*}SQ^2 &= (x_2-x_1)^2 + (y_2-y_1)^2 \\ SQ^2 &= (\cos \alpha - \cos \beta)^2 + (\sin \alpha - \sin \beta)^2 \\ &= (\cos ^2 \alpha - 2\cos \alpha \cos \beta + \cos ^2 \beta) + (\sin ^2 \alpha - 2\sin \alpha \sin \beta + \sin ^2 \beta) \\ &= ( \sin ^2 \alpha + \cos ^2 \alpha ) + (\sin ^2 \beta + \cos ^2 \beta ) - 2(\cos \alpha \cos \beta + \sin \alpha \sin b) \\ &= (1) + (1 ) - 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta) \\ SQ^2 &= 2 - 2(\cos \alpha \cos b + \sin a \sin b) \end{align*}$$ **Cosine rule in triangle POQ**
Note that $\triangle SOQ$ applies cosinus $SQ^2 = OS^2 + OQ^2 - 2.OS.OQ .\cos (\alpha-\beta)$ and then substitusi $SQ^2 = 2 - 2(\ cos \alpha \cos \beta + \sin \alpha \sin \beta)$ $$\begin{align*} SQ^2 &= OS^2 + OQ^2 - 2.OS.OQ .\cos (\alpha-\beta)\\ SQ^2 &= 1^2 + 1 ^2 - 2.1.1 . \cos (\alpha-\beta) \\ 2 - 2 \cos (\alpha-\beta) &= SQ^2 \\ & \text{(substitusi } SQ^2 ) \\ 2 - 2 \cos (\alpha-\beta) &= 2 - 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta)\\ \cos (\alpha-\beta) &= \cos \alpha \cos \beta + \sin \alpha \sin \beta \end{align*}$$
so it is proven: $\cos (\alpha-\beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$

Example of a Trigonmetry Problem on the Sum and Difference of Two Cosine Angles

1. Determine the value of $\cos 15^\circ$
   

   Solution ✍️

   
   Multiply out $\cos (a-b) = \cos a \cos b + \sin a \sin b$. $$\begin{align*} \cos 15^\circ & = \cos (45^\circ - 30^\circ) \\ & = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ \\ & = \frac{1}{2}\sqrt{2} . \frac{1}{2}\sqrt{3} + \frac{1}{2}\sqrt{2} . \frac{1}{2} \\ & = \frac{1}{4}\sqrt{6}+ \frac{1}{4}\sqrt{2} \\ & = \frac{1}{4}\left( \sqrt{6} +\sqrt{2}\right ) \\ & = \frac{1}{4}\sqrt{2} (\frac{1}{2}\sqrt{3} + 1) \end{align*}$$ So, the value of $\cos 105^\circ$ is $\frac{1}{4}\sqrt{2} (\frac{1}{2}\sqrt{3} + 1)$.

2. Determine the value of $\cos 105^\circ$
   

   Solution ✍️

   
   Multiply out $\cos (a+b) = \cos a \cos b - \sin a \sin b$. $$\begin{align*} \cos 105^\circ & = \cos (60^\circ + 45^\circ) \\ & = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ \\ & = \frac{1}{2}. \frac{1}{2}\sqrt{2} - \frac{1}{2}\sqrt{3} . \frac{1}{2}\sqrt{2} \\ & = \frac{1}{4}\sqrt{2}- \frac{1}{4}\sqrt{6} \\ & = \frac{1}{4}\left( \sqrt{2} -\sqrt{6}\right ) \\ & = \frac{1}{4}\sqrt{2} (1-\frac{1}{2}\sqrt{3}) \end{align*}$$

3. Determine the value of $\cos 88^\circ \cos 58^\circ + \sin 88^\circ \sin 58^\circ$
   

   Solution ✍️

   
   Multiply out $\cos a \cos b + \sin a \sin b=\cos (a-b)$. $$\begin{align*} \cos 88^\circ \cos 58^\circ + \sin 88^\circ \sin 58^\circ & = \cos (88^\circ-58^\circ)\\ &= \cos 30^\circ \\ &= \frac{1}{2}\sqrt{3}\end{align*}$$ So, the value of $\cos 88^\circ \cos 58^\circ + \sin 88^\circ \sin 58^\circ$ is $\frac{1}{2}\sqrt{3}$

4. Known Known $\cos x =\dfrac35$ and $\cos y = \dfrac{12}{13}$ ($x$ and $y$ are acute angles). Determine the value of $\cos(x-y)$!
   

   Solution ✍️

   
   Gunakan rumus $\cos (x-y)=\cos x \cos y + \sin x \sin y$.
   cos $x$ and cos $y$ are already known, so we need to determine $\sin x$ and $\sin y$ first using the identity formula $\sin^2 x+\cos^2x=1$ or you can also use draw a triangle.

From the Identity $\sin^2 x+\cos^2x=1$,

• so $\sin^2 x=1-\cos^2x$. $$\begin{align*}\sin^2 x&=1-\cos^2x\\ \sin x&=+\sqrt{1-\cos^2x} \text{ … }x \text { thin, so } \sin x \text{ positif}\\ &=+\sqrt{1-\left(\frac35\right)^2}\\ &=+\sqrt{1-\frac{9}{25}}\\ &=\sqrt{\frac{25}{25}-\frac{9}{25}}=\sqrt{\frac{16}{25}}\\ \sin x&=\frac{4}{5}\end{align*}$$

• next $\sin^2 y=1-\cos^2y$. $$\begin{align*}\sin^2 y&=1-\cos^2y\\ \sin y&=+\sqrt{1-\cos^2y} \text{ … }y \text { thin, so } \sin y \text{ positif }\\ &=+\sqrt{1-\left(\frac{12}{13}\right)^2}\\ &=+\sqrt{1-\frac{144}{169}}\\ &=\sqrt{\frac{169}{169}-\frac{144}{169}}=\sqrt{\frac {25}{169}}\\ \sin y&=\frac{5}{13}\end{align*}$$

• calculates the value of $\cos (x-y)$ $$\begin{align*} \cos (x-y)&=\cos x \cos y + \sin x \sin y \\ &=\left(\frac{3}{5}\right)\left(\frac{12}{13}\right) + \left(\frac{4}{5}\right)\left(\frac{5}{13}\right) \\ &=\frac{36}{65} + \frac{20}{65} \\ &=\frac{56}{65} \end{align*}$$ So, the value of $\cos(x-y)=\dfrac{56}{65}$

2. Formulas for $\sin(\alpha+\beta)$ and $\sin(\alpha-\beta)$

To prove the formulas $sin(\alpha+\beta)$ and $sin(\alpha-\beta)$ above, we simply use the properties of the angle relations in quadrant I from the proof of $cos(\alpha+\beta)$ above. Actually there is another way, namely by using the area of ​​the triangle. However, this time we only use angle relations in quadrant I.

Remember the angle relation $\sin A=\cos (90^\circ-A)$

Proof formula $\sin(\alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

$$\begin{align*} \sin ( \alpha + \beta ) &= \cos [90^\circ - ( \alpha + \beta )] \\ &= \cos [90^\circ - \alpha - \beta ] \\ &= \cos [(90^\circ - \alpha) - \beta ] \\ &= \cos (90^\circ - \alpha) \cos \beta + \sin (90^\circ - \alpha) \sin \beta \\ \sin ( \alpha + \beta ) &= \sin \alpha \cos \beta + \cos \alpha \sin \beta \end{align*}$$ Jadi, prove : $\sin ( \alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

Proof formula $\sin(\alpha - \beta ) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$

$$\begin{align*} \sin ( \alpha - \beta ) &= \sin ( \alpha +(-\beta))\\ &= \sin \alpha \cos (-\beta ) + \ cos \alpha \sin ( - \beta ) \\ &= \sin \alpha \cos \beta + \cos \alpha . (- \sin \beta ) \\ \sin ( \alpha - \beta ) &= \sin \alpha \cos \beta - \cos \alpha \sin \beta \end{align*}$$ Jadi, prove : $\sin ( \alpha - \beta ) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$

Example of a Trigonmetry Problem on the Sum and Difference of Two Sine Angles

1). Determine the value of $\sin 15^\circ$

Solution ✍️

Multiply out $\sin (a-b) = \sin a \cos b + \cos a \sin b$. $$\begin{align*} \sin 15^\circ & = \sin (45^\circ - 30^\circ) \\ & = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \\ & = \frac{1}{2}\sqrt{2} . \frac{1}{2}\sqrt{3} + \frac{1}{2}\sqrt{2} . \frac{1}{2} \\ & = \frac{1}{4}\sqrt{6}- \frac{1}{4}\sqrt{2} \\ & = \frac{1}{4}\left( \sqrt{6} -\sqrt{2}\right ) \\ & = \frac{1}{4}\sqrt{2} (\frac{1}{2}\sqrt{3} - 1) \end{align*}$$ So, the value of $\sin 15^\circ$ is $\frac{1}{4}\sqrt{2} (\frac{1}{2}\sqrt{3} - 1)$

2). Determine the value of $\sin 75^\circ$

Solution ✍️

Multiply out $\sin (a+b) = \sin a \cos b + \cos a \sin b$. $$\begin{align*} \sin 75^\circ & = \sin (45^\circ + 30^\circ) \\ & = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \\ & = \frac{1}{2}\sqrt{2}. \frac{1}{2}\sqrt{3} + \frac{1}{2}\sqrt{2} . \frac{1}{2} \\ & = \frac{1}{4}\sqrt{6}+ \frac{1}{4}\sqrt{2} \\ & = \frac{1}{4}\left( \sqrt{6} +\sqrt{2}\right ) \\ & = \frac{1}{4}\sqrt{2} (\frac{1}{2}\sqrt{3}+1) \end{align*}$$ So, the value of $\sin 75^\circ$ is $\frac{1}{4}\sqrt{2} (\frac{1}{2}\sqrt{3}+1)$

3). Determine the value of $\sin 64^\circ \cos 56^\circ + \cos 64^\circ \sin 41^\circ$

Solution ✍️

Multiply out $\sin a \cos b + \cos a \sin b=\sin (a+b)$. $$\begin{align*} \sin 64^\circ \cos 56^\circ + \cos 64^\circ \sin 56^\circ & = \sin (64^\circ+56^\circ)\ \ &= \sin 120^\circ=\sin 30^\circ \\ &= \frac{1}{2}\end{align*}$$ So, the value of the value of $\sin 64^\circ \cos 56^\circ + \cos 64^\circ \sin 41^\circ$ is $\dfrac{1}{2}$.

4). Given Given $\sin p =\dfrac35$ and $\cos q = \dfrac{12}{13}$ ($p$ in quadrant III and $q$ acute angle). Determine the value of $\sin(p+q)$!

Solution ✍️

The multiplier is $\sin (p+q)=\sin p \cos q + \cos p \sin q$.
sin $p$ and cos $q$ are already known, so we need to determine $\cos p$ and $\sin q$ first using the identity formula $\sin^2 p+\cos^2q=1$ or you can also use draw a triangle.
From the Identity $\sin^2 p+\cos^2q=1$,

*) so $\cos^2 p=1-\sin^2q$.

$$\begin{align*}\cos^2 p&=1-\sin^2p\\ \cos p&=-\sqrt{1-\sin^2p} \text{ … }p \text { in quadrant III, so } \cos p \text{ negatif}\\ &=-\sqrt{1-\left(\frac35\right)^2}\\ &=-\sqrt{1-\frac{9}{25}}\\ &=-\sqrt{\frac{25}{25}-\frac{9}{25}}=-\sqrt{\frac{16}{25}}\\ \cos p&=-\frac{4}{5}\end{align*}$$

*) next $\sin^2 q=1-\cos^2q$.

$$\begin{align*}\sin^2 q&=1-\cos^2q\\ \sin q&=+\sqrt{1-\cos^2q} \text{ … }q \text { thin, so } \sin q \text{ positif}\\ &=+\sqrt{1-\left(\frac{12}{13}\right)^2}\\ &=+\sqrt{1-\frac{144}{169}}\\ &=\sqrt{\frac{169}{169}-\frac{144}{169}}=\sqrt{\frac{25}{169}}\\ \sin q&=\frac{5}{13}\end{align*}$$

*) calculates the value of $\sin (p+q)$

$$\begin{align*} \sin (x+y)&=\sin p \cos q + \cos p \sin q \\ &=\left(\frac{3}{5}\right)\left(\frac{12}{13}\right) + \left(-\frac{4}{5}\right)\left(\frac{5}{13}\right) \\ &=\frac{36}{65} - \frac{20}{65} \\ &=\frac{16}{65} \end{align*}$$ So, the value of $\sin(p+q)=\dfrac{16}{65}$

3. Formulas for $\tan(\alpha+\beta)$ and $\tan(\alpha-\beta)$

To prove the formulas $\tan(\alpha+\beta)$ and $\tan(\alpha-\beta)$ above we use the trigonometry formula for the sum of two angles $cos(\alpha+\beta)$ and $cos(\alpha+\beta )$. Apart from that, we also use the trigonometric identity formula $\tan A=\dfrac{\sin A}{\cos B}$.

Proof of formula $\tan ( \alpha + \beta ) = \dfrac{\tan \alpha + \beta}{1 - \tan \tan \beta }$

$$\begin{align*} \tan ( \alpha + \beta ) & = \frac{ \sin ( \alpha + \beta ) }{\cos ( \alpha + \beta )} \\ & = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta} \\ & = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta} . \frac{\frac{1}{\cos \alpha \cos \beta}}{\frac{1}{\cos \alpha \cos \beta}} \\ & = \frac{\frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta}}{\frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\cos \alpha \cos \beta}} \\ & = \frac{\frac{\sin \alpha \cos \beta }{\cos \alpha \cos \beta} + \frac{\cos \alpha \sin \beta}{\cos \alpha \cos \beta}}{\frac{ \cos \alpha \cos \beta}{\cos \alpha \cos \beta} - \frac{ \sin \alpha \sin \beta}{\cos \alpha \cos \beta}} \\ & = \frac{\frac{\sin \alpha }{\cos \alpha } + \frac{ \sin \beta}{ \cos \beta}}{1 - \frac{ \sin \alpha }{\cos \alpha }\frac{ \sin \beta}{ \cos \beta}} \\ \tan ( \alpha + \beta ) & = \frac{ \tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta } \end{align*}$$

It is proven that $\tan ( \alpha + \beta ) = \dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta }$

Proof of formula $\tan ( \alpha - \beta ) = \dfrac{\tan \alpha - \tan \beta}{1 + \tan \tan \beta }$

To prove this we use related angles in quadrant IV or negative angles. Remember that $\tan (-A)=-\tan A$. $$\begin{align*} \tan ( \alpha - \beta ) & = \tan ( \alpha + (- \beta )) \\ & = \frac{ \tan \alpha + \tan (-\beta )}{1 - \tan \alpha \tan (-\beta ) } \\ & = \frac{ \tan \alpha - \tan \beta }{1 - \tan \alpha . (- \tan \beta ) } \\ & = \frac{ \tan \alpha - \tan \beta }{1 + \tan \alpha \tan \beta } \end{align*}$$

Example Trigonometry Question Sum and Difference of Two Tangent Angles

1). Without using calculators and trigonometry tables. Determine the value of $\tan 75^\circ$!

Solution ✍️

Use the formula $\tan (a+b) = \dfrac{\tan a + \tan b}{1-\tan a. \tan b}$. $$\begin{align*} \tan 75^\circ & = \tan ( 45^\circ + 30^\circ ) \\ & = \frac{ \tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ } \\ & = \frac{ 1 + \frac{1}{3} \sqrt{3} }{1 - 1.\frac{1}{3} \sqrt{3} } \\ & = \frac{ 1 + \frac{1}{3} \sqrt{3} }{1 - \frac{1}{3} \sqrt{3} } \times \frac{3}{3} \\ & = \frac{ 3 + \sqrt{3} }{3 - \sqrt{3} } \\ & = \frac{ 3 + \sqrt{3} }{3 - \sqrt{3} } \times \frac{ 3 + \sqrt{3} }{3 + \sqrt{3} } \\ & = \frac{ 9 + 6\sqrt{3} + 3 }{9 - 3 }\\& = \frac{ 12 + 6\sqrt{3} }{6 }\\ \tan 75^\circ & = 2 + \sqrt{3} \end{align*}$$ So, the value of $\tan 75^\circ= 2+\sqrt{3}$.

2). Determine the value of $\tan 255^\circ$

Solution ✍️

Use the formula $\tan (a-b) = \dfrac{\tan a - \tan b}{1+\tan a. \tan b}$. $$\begin{align*} \tan 255^\circ &= \tan(300^\circ-45^\circ)\\ &= \frac{\tan 300^\circ - \tan 45^\circ}{1+\tan 300^\circ \tan 45^\circ}\\ &= \frac{-\tan 60^\circ - \tan 45^\circ}{1-\tan 60^\circ \tan 45^\circ}\\ &= \frac{-\sqrt{3} - 1}{1-\sqrt{3} \cdot 1}\\ &= \frac{-\sqrt{3}-1}{-\sqrt{3}+1}\times \frac{-1}{-1}\\ &= \frac{\sqrt{3}+1}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}\\ &= \frac{3+2\sqrt{3}+1}{3-1}\\ &= \frac{4+2\sqrt{3}}{2}\\ &= 2+\sqrt{3} \end{align*}$$ So, the value of $\tan 255^\circ= 2+\sqrt{3}$.

3). Given: $\sin p =\dfrac35$ and $\cos q = \dfrac{12}{13}$ ($p$ in quadrant I and $q$ in quadrant IV). Determine the value of $\tan(p+q)$!

Solution ✍️

Use the formula $\tan (a+b) = \dfrac{\tan a+\tan b}{1-\tan a. \tan b}$.
sin $p$ and cos $q$ are already known, so we need to determine $\tan p$ and $\tan q$ first by just drawing a triangle.

Remember trigonometric ratios
$\sin p =\dfrac{opposite}{hypotenuse}=\dfrac{de}{mi}$
$\cos p =\dfrac{adjacent}{hypotenuse}=\dfrac{sa}{mi}$
$\tan p =\dfrac{opposite}{adjacent}=\dfrac{de}{sa}$
For $\sin p =\dfrac35=\dfrac{de}{mi}$ then front 3 slopes 5 and the value of $\tan p$ is positive because it is in quadrant I.
For $\cos p =\dfrac{12}{13}=\dfrac{sa}{mi}$ then front 3 slopes 5 and the value of $\tan q$ is negative because it is in quadrant IV.
Look at the pictures!

*) find the value $\tan p$ but first we need to find the side length of the triangle with angle $p$ using Pythagoras.

$$sa=\sqrt{5^2-3^2}=\sqrt{25-9}=\sqrt{16}=4$$ So that the value of $\tan p=\dfrac{de}{sa}=\dfrac{3}{4}$.

*) find the value of $\tan q$ but first we need to find the side length of the triangle with angle $q$ using Pythagoras.

$$de=\sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5$$ So the value of $\tan q=-\dfrac{de}{sa}=-\dfrac{5}{12}$ (negative because it is in quadrant IV).

*) find the value $\tan (p+q)$.

$$\begin{align*} \tan (p+q) &= \frac{\tan p+\tan q}{1-\tan p. \tan q}\\ &=\frac{\frac{3}{4}+\left(-\frac{5}{12}\right)}{1-\frac{3}{4}.\left(-\frac{5}{12}\right)}\\ &=\frac{\frac{3}{4}-\frac{5}{12}}{1+\frac{15}{48}}\\ &=\frac{\frac{36}{48}-\frac{20}{48}}{\frac{48}{48}+\frac{15}{48}}\\ &=\frac{\frac{16}{48}}{\frac{63}{48}}\\ &=\frac{16}{63} \end{align*}$$ So, the value of $\tan(p+q)=\dfrac{16}{63}$