Question

Prove the trigonometric expression: \[\sin {{55}^{\circ }}\text{sin}{{35}^{\circ }}-\cos {{55}^{\circ }}\cos {{35}^{\circ }}\text{=}0\].

Answer 1

Hint: Use the identity $\cos (A+B)=\cos A\cos B-\sin A\sin B$ and then substitute ‘A’ as ‘55’ and ‘B’ as ‘35’. Then use the fact related to 90 and get the result.

Complete step-by-step answer:

In the question we are asked to prove that \[\sin {{55}^{\circ }}\text{sin}{{35}^{\circ }}-\cos {{55}^{\circ }}\cos {{35}^{\circ }}\text{=}0\].
So for this we will apply the formula of $\cos (A+B)=\cos A\cos B-\sin A\sin B$.
Here if we substitute the value of A = 55 and B = 35, so the above formula can be re-written as,
$\cos ({{55}^{\circ }}+{{35}^{\circ }})=\cos {{55}^{\circ }}\cos {{35}^{\circ }}-\sin {{55}^{\circ }}\sin {{35}^{\circ }}$
Now adding the values in left hand side, we get
$\cos ({{90}^{\circ }})=\cos {{55}^{\circ }}\cos {{35}^{\circ }}-\sin {{55}^{\circ }}\sin {{35}^{\circ }}$
We know, $\cos {{90}^{\circ }}=0$, so the above expression can be written as,
$0=\cos {{55}^{\circ }}\cos {{35}^{\circ }}-\sin {{55}^{\circ }}\sin {{35}^{\circ }}$
Now we can multiply by (-1) throughout the equation, the above expression can be written as
$0\times (-1)=-1\left( \cos {{55}^{\circ }}\cos {{35}^{\circ }}-\sin {{55}^{\circ }}\sin {{35}^{\circ }} \right)$
Simplifying the above equation, we can rewrite it as
$0=\left( \sin {{55}^{\circ }}\sin {{35}^{\circ }}-\cos {{55}^{\circ }}\cos {{35}^{\circ }} \right)$
This is the same as the expression given in the question.
Hence, the given trigonometric expression is proved.

Note: We can solve it in another way by using $\cos \left( 90-\theta \right)=\sin \theta $. So we can change both $cos{{55}^{\circ }},\cos {{35}^{\circ }}$ into their respective sine ratios which are $sin{{35}^{\circ }},sin{{55}^{\circ }}$respectively. After that we can see that the expression changes from $\left( \sin {{55}^{\circ }}\sin {{35}^{\circ }}-\cos {{55}^{\circ }}\cos {{35}^{\circ }} \right)$to $\left( \sin {{55}^{\circ }}\sin {{35}^{\circ }}-\sin {{55}^{\circ }}\sin {{35}^{\circ }} \right)$, and this is equal to zero. In this way also we can prove the given expression too.
Student often goes wrong in writing the formula, $\cos (A+B)=\cos A\cos B-\sin A\sin B$
Due to which they will get the wrong answer.