Question

How do you use the double angle or half angle formula to simplify $2\sin 35{}^\circ \cos 35{}^\circ $.

Answer 1

Hint: To solve the above question we will use the concept of trigonometric identities. We will use the formula $\sin 2\theta =2\sin \theta \cos \theta $ and $2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right)$ to solve the above question. When we use the formula $\sin 2\theta =2\sin \theta \cos \theta $, then $\theta =35{}^\circ $ which we get after comparing from equation $2\sin 35{}^\circ \cos 35{}^\circ $. So, we can write the above equation as $\sin \left( 2\times 35{}^\circ \right)=2\sin 35{}^\circ \cos 35{}^\circ $
$\Rightarrow \sin \left( 2\times 35{}^\circ \right)=2\sin 35{}^\circ \cos 35{}^\circ $
$\Rightarrow 2\sin 35{}^\circ \cos 35{}^\circ =\sin 70{}^\circ $

Complete step-by-step solution:
We will use the concept of the trigonometric identities to solve the above question. We will use the formula $\sin 2\theta =2\sin \theta \cos \theta $ and $2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right)$ to solve the above question.
We will compare $2\sin \theta \cos \theta $ with the $2\sin 35{}^\circ \cos 35{}^\circ $, then we will get $\theta =35{}^\circ $.
Since, we know that $\sin 2\theta =2\sin \theta \cos \theta $, so when we put $\theta =35{}^\circ $, we will get:
$\Rightarrow \sin 2\theta =2\sin \theta \cos \theta $
So, when $\theta =35{}^\circ $ we will get:
$\Rightarrow \sin 2\times 35{}^\circ =2\sin 35{}^\circ \cos 35{}^\circ $
$\Rightarrow 2\sin 35{}^\circ \cos 35{}^\circ =\sin 70{}^\circ $
So, we can say that the simplified form of $2\sin 35{}^\circ \cos 35{}^\circ $is equal to $\sin 70{}^\circ $.
Now, we can also solve the above question alternatively using the formula $2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right)$.
When we compare $2\sin 35{}^\circ \cos 35{}^\circ $with $2\sin A\cos B$, then we will get:
$A=35{}^\circ ,B=35{}^\circ $
Now, we know that $2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right)$. So, after putting the value of A and B in the above equation we will get:
$\Rightarrow 2\sin 35{}^\circ \cos 35{}^\circ =\sin \left( 35{}^\circ +35{}^\circ \right)+\sin \left( 35{}^\circ -35{}^\circ \right)$
Now, after simplifying we will get:
$\Rightarrow 2\sin 35{}^\circ \cos 35{}^\circ =\sin 70{}^\circ +\sin 0{}^\circ $
Now, we know that value of $\sin 0{}^\circ =0$.
So, the value of $2\sin 35{}^\circ \cos 35{}^\circ =\sin 70{}^\circ +0$
$\Rightarrow 2\sin 35{}^\circ \cos 35{}^\circ =\sin 70{}^\circ $
This is our required solution.

Note: Students are required to note that when they are solving trigonometry questions then they must revise and first memorize the trigonometric formulas otherwise they will not be able to solve the question especially when they have to prove or simplify any trigonometric expression.