
What are some sum and difference identities examples?
Answer
528.3k+ views
Hint: For solving this question you can use the sum and difference identities of trigonometry. These identities are for $\sin ,\cos ,\tan $. And we can solve any type of question of trigonometric sum of angles and differences of angles easily. And these will always be in the form of angle $\left( a+b \right)$ for sum and $\left( a-b \right)$ for difference.
Complete step by step solution:
According to our question we have to give some examples of sum and difference identities of trigonometric functions. So, if we see the sum and difference identities, then they are as following:
$\begin{align}
& \sin \left( a+b \right)=\sin a\cos b+\cos a\sin b \\
& \sin \left( a-b \right)=\sin a\cos b-\cos a\sin b \\
& \cos \left( a+b \right)=\cos a\cos b-\sin a\sin b \\
& \cos \left( a-b \right)=\cos a\cos b+\sin a\sin b \\
& \tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b} \\
& \tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a\tan b} \\
\end{align}$
These are the identities of sum and difference of trigonometric functions. If we take an example for these.
Example 1. Find $\cos \left( \dfrac{5\pi }{12} \right)$
We can write it as $\cos \left( \dfrac{5\pi }{12} \right)=\cos \left( \dfrac{\pi }{6}+\dfrac{\pi }{4} \right)$
Which is a form of $\cos \left( a+b \right)$ and it is given as,
$\cos \left( a+b \right)=\cos a\cos b-\sin a\sin b$
So, we will write the expression as,
$\begin{align}
& \cos \left( \dfrac{\pi }{4}+\dfrac{\pi }{6} \right)=\cos \dfrac{\pi }{4}\cos \dfrac{\pi }{6}-\sin \dfrac{\pi }{4}\sin \dfrac{\pi }{6} \\
& \Rightarrow \cos \left( \dfrac{\pi }{4}+\dfrac{\pi }{6} \right)=\dfrac{\sqrt{2}}{2}.\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{2}.\dfrac{1}{2} \\
& \Rightarrow \cos \left( \dfrac{\pi }{4}+\dfrac{\pi }{6} \right)=\dfrac{\sqrt{6}}{4}-\dfrac{\sqrt{2}}{4} \\
& \Rightarrow \cos \left( \dfrac{\pi }{4}+\dfrac{\pi }{6} \right)=\dfrac{\sqrt{6}-\sqrt{2}}{4} \\
\end{align}$
Example 2. Find $\cos \left( {{42}^{\circ }} \right)\cos \left( {{18}^{\circ }} \right)-\sin \left( {{42}^{\circ }} \right)\sin \left( {{18}^{\circ }} \right)$ exactly.
It is like $\cos a\cos b-\sin a\sin b$ form and we know that $\cos a\cos b-\sin a\sin b=\cos \left( a+b \right)$. So, we can write the given expression as follows,
$\begin{align}
& \cos \left( {{42}^{\circ }} \right)\cos \left( {{18}^{\circ }} \right)-\sin \left( {{42}^{\circ }} \right)\sin \left( {{18}^{\circ }} \right)=\cos \left( {{42}^{\circ }}+{{18}^{\circ }} \right) \\
& \Rightarrow \cos \left( {{42}^{\circ }} \right)\cos \left( {{18}^{\circ }} \right)-\sin \left( {{42}^{\circ }} \right)\sin \left( {{18}^{\circ }} \right)=\cos \left( {{60}^{\circ }} \right) \\
& \Rightarrow \cos \left( {{42}^{\circ }} \right)\cos \left( {{18}^{\circ }} \right)-\sin \left( {{42}^{\circ }} \right)\sin \left( {{18}^{\circ }} \right)=\dfrac{1}{2} \\
\end{align}$
Example 3. Find $\dfrac{\tan \left( {{80}^{\circ }} \right)-\tan \left( {{35}^{\circ }} \right)}{1+\tan \left( {{80}^{\circ }} \right)\tan \left( {{35}^{\circ }} \right)}$ exactly.
If we compare $\dfrac{\tan \left( {{80}^{\circ }} \right)-\tan \left( {{35}^{\circ }} \right)}{1+\tan \left( {{80}^{\circ }} \right)\tan \left( {{35}^{\circ }} \right)}$ as $\dfrac{\tan a-\tan b}{\tan a\tan b+1}$, then from the identities of sum and differences we can write it as $\tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a\tan b}$. So, we will get as follows,
$\begin{align}
& \tan \left( {{80}^{\circ }}-{{35}^{\circ }} \right)=\dfrac{\tan \left( {{80}^{\circ }} \right)-\tan \left( {{35}^{\circ }} \right)}{1+\tan \left( {{80}^{\circ }} \right)\tan \left( {{35}^{\circ }} \right)} \\
& \Rightarrow \dfrac{\tan \left( {{80}^{\circ }} \right)-\tan \left( {{35}^{\circ }} \right)}{1+\tan \left( {{80}^{\circ }} \right)\tan \left( {{35}^{\circ }} \right)}=\tan \left( {{45}^{\circ }} \right) \\
& \Rightarrow \dfrac{\tan \left( {{80}^{\circ }} \right)-\tan \left( {{35}^{\circ }} \right)}{1+\tan \left( {{80}^{\circ }} \right)\tan \left( {{35}^{\circ }} \right)}=1 \\
\end{align}$
So, these are some of the examples of sum and difference identities.
Note: During solving this type of questions, we should be careful about converting them from a form to another one. It is done by using the formula but we should be careful of the value of angles and before that we have to find an identity for it for its proper form and then we can change these with one another.
Complete step by step solution:
According to our question we have to give some examples of sum and difference identities of trigonometric functions. So, if we see the sum and difference identities, then they are as following:
$\begin{align}
& \sin \left( a+b \right)=\sin a\cos b+\cos a\sin b \\
& \sin \left( a-b \right)=\sin a\cos b-\cos a\sin b \\
& \cos \left( a+b \right)=\cos a\cos b-\sin a\sin b \\
& \cos \left( a-b \right)=\cos a\cos b+\sin a\sin b \\
& \tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b} \\
& \tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a\tan b} \\
\end{align}$
These are the identities of sum and difference of trigonometric functions. If we take an example for these.
Example 1. Find $\cos \left( \dfrac{5\pi }{12} \right)$
We can write it as $\cos \left( \dfrac{5\pi }{12} \right)=\cos \left( \dfrac{\pi }{6}+\dfrac{\pi }{4} \right)$
Which is a form of $\cos \left( a+b \right)$ and it is given as,
$\cos \left( a+b \right)=\cos a\cos b-\sin a\sin b$
So, we will write the expression as,
$\begin{align}
& \cos \left( \dfrac{\pi }{4}+\dfrac{\pi }{6} \right)=\cos \dfrac{\pi }{4}\cos \dfrac{\pi }{6}-\sin \dfrac{\pi }{4}\sin \dfrac{\pi }{6} \\
& \Rightarrow \cos \left( \dfrac{\pi }{4}+\dfrac{\pi }{6} \right)=\dfrac{\sqrt{2}}{2}.\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{2}.\dfrac{1}{2} \\
& \Rightarrow \cos \left( \dfrac{\pi }{4}+\dfrac{\pi }{6} \right)=\dfrac{\sqrt{6}}{4}-\dfrac{\sqrt{2}}{4} \\
& \Rightarrow \cos \left( \dfrac{\pi }{4}+\dfrac{\pi }{6} \right)=\dfrac{\sqrt{6}-\sqrt{2}}{4} \\
\end{align}$
Example 2. Find $\cos \left( {{42}^{\circ }} \right)\cos \left( {{18}^{\circ }} \right)-\sin \left( {{42}^{\circ }} \right)\sin \left( {{18}^{\circ }} \right)$ exactly.
It is like $\cos a\cos b-\sin a\sin b$ form and we know that $\cos a\cos b-\sin a\sin b=\cos \left( a+b \right)$. So, we can write the given expression as follows,
$\begin{align}
& \cos \left( {{42}^{\circ }} \right)\cos \left( {{18}^{\circ }} \right)-\sin \left( {{42}^{\circ }} \right)\sin \left( {{18}^{\circ }} \right)=\cos \left( {{42}^{\circ }}+{{18}^{\circ }} \right) \\
& \Rightarrow \cos \left( {{42}^{\circ }} \right)\cos \left( {{18}^{\circ }} \right)-\sin \left( {{42}^{\circ }} \right)\sin \left( {{18}^{\circ }} \right)=\cos \left( {{60}^{\circ }} \right) \\
& \Rightarrow \cos \left( {{42}^{\circ }} \right)\cos \left( {{18}^{\circ }} \right)-\sin \left( {{42}^{\circ }} \right)\sin \left( {{18}^{\circ }} \right)=\dfrac{1}{2} \\
\end{align}$
Example 3. Find $\dfrac{\tan \left( {{80}^{\circ }} \right)-\tan \left( {{35}^{\circ }} \right)}{1+\tan \left( {{80}^{\circ }} \right)\tan \left( {{35}^{\circ }} \right)}$ exactly.
If we compare $\dfrac{\tan \left( {{80}^{\circ }} \right)-\tan \left( {{35}^{\circ }} \right)}{1+\tan \left( {{80}^{\circ }} \right)\tan \left( {{35}^{\circ }} \right)}$ as $\dfrac{\tan a-\tan b}{\tan a\tan b+1}$, then from the identities of sum and differences we can write it as $\tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a\tan b}$. So, we will get as follows,
$\begin{align}
& \tan \left( {{80}^{\circ }}-{{35}^{\circ }} \right)=\dfrac{\tan \left( {{80}^{\circ }} \right)-\tan \left( {{35}^{\circ }} \right)}{1+\tan \left( {{80}^{\circ }} \right)\tan \left( {{35}^{\circ }} \right)} \\
& \Rightarrow \dfrac{\tan \left( {{80}^{\circ }} \right)-\tan \left( {{35}^{\circ }} \right)}{1+\tan \left( {{80}^{\circ }} \right)\tan \left( {{35}^{\circ }} \right)}=\tan \left( {{45}^{\circ }} \right) \\
& \Rightarrow \dfrac{\tan \left( {{80}^{\circ }} \right)-\tan \left( {{35}^{\circ }} \right)}{1+\tan \left( {{80}^{\circ }} \right)\tan \left( {{35}^{\circ }} \right)}=1 \\
\end{align}$
So, these are some of the examples of sum and difference identities.
Note: During solving this type of questions, we should be careful about converting them from a form to another one. It is done by using the formula but we should be careful of the value of angles and before that we have to find an identity for it for its proper form and then we can change these with one another.
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