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Question: Prove the trigonometric expression: \[\sin {{55}^{\circ }}\text{sin}{{35}^{\circ }}-\cos {{55}^{\cir...

Prove the trigonometric expression: sin⁡55∘sin35∘−cos⁡55∘cos⁡35∘=0\sin {{55}^{\circ }}\text{sin}{{35}^{\circ }}-\cos {{55}^{\circ }}\cos {{35}^{\circ }}\text{=}0.

Explanation

Solution

Hint: Use the identity cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\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∘sin35∘−cos⁡55∘cos⁡35∘=0\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⁡Acos⁡B−sin⁡Asin⁡B\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∘+35∘)=cos⁡55∘cos⁡35∘−sin⁡55∘sin⁡35∘\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∘)=cos⁡55∘cos⁡35∘−sin⁡55∘sin⁡35∘\cos ({{90}^{\circ }})=\cos {{55}^{\circ }}\cos {{35}^{\circ }}-\sin {{55}^{\circ }}\sin {{35}^{\circ }}
We know, cos⁡90∘=0\cos {{90}^{\circ }}=0, so the above expression can be written as,
0=cos⁡55∘cos⁡35∘−sin⁡55∘sin⁡35∘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×(−1)=−1(cos⁡55∘cos⁡35∘−sin⁡55∘sin⁡35∘)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=(sin⁡55∘sin⁡35∘−cos⁡55∘cos⁡35∘)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⁡(90−θ)=sin⁡θ\cos \left( 90-\theta \right)=\sin \theta . So we can change both cos55∘,cos⁡35∘cos{{55}^{\circ }},\cos {{35}^{\circ }} into their respective sine ratios which are sin35∘,sin55∘sin{{35}^{\circ }},sin{{55}^{\circ }}respectively. After that we can see that the expression changes from (sin⁡55∘sin⁡35∘−cos⁡55∘cos⁡35∘)\left( \sin {{55}^{\circ }}\sin {{35}^{\circ }}-\cos {{55}^{\circ }}\cos {{35}^{\circ }} \right)to (sin⁡55∘sin⁡35∘−sin⁡55∘sin⁡35∘)\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⁡Acos⁡B−sin⁡Asin⁡B\cos (A+B)=\cos A\cos B-\sin A\sin B
Due to which they will get the wrong answer.