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Question: How do you find the exact value of \(\sin 33{}^\circ \cos 27{}^\circ +\cos 33{}^\circ \sin 27{}^\cir...

How do you find the exact value of sin⁡33∘cos⁡27∘+cos⁡33∘sin⁡27∘\sin 33{}^\circ \cos 27{}^\circ +\cos 33{}^\circ \sin 27{}^\circ .

Explanation

Solution

To solve the above question we will use the concept of trigonometric identities. We will use the formula sin⁡(A+B)=sin⁡Acos⁡B+sin⁡Bcos⁡A\sin \left( A+B \right)=\sin A\cos B+\sin B\cos A to solve the above question. At first, we will compare the equation given in the question sin⁡33∘cos⁡27∘+cos⁡33∘sin⁡27∘\sin 33{}^\circ \cos 27{}^\circ +\cos 33{}^\circ \sin 27{}^\circ with sin⁡Acos⁡B+sin⁡Bcos⁡A\sin A\cos B+\sin B\cos A to get the value of A and B. Then, we will put the value of A and B in equation sin⁡(A+B)=sin⁡Acos⁡B+sin⁡Bcos⁡A\sin \left( A+B \right)=\sin A\cos B+\sin B\cos A and get the required answer.

Complete step-by-step solution:
We can see that the above given question is of trigonometry, so we will use trigonometric identity to solve the above question.
Since, we have to find the value of sin⁡33∘cos⁡27∘+cos⁡33∘sin⁡27∘\sin 33{}^\circ \cos 27{}^\circ +\cos 33{}^\circ \sin 27{}^\circ .
We can see that the above expression is similar to the sum of two different angle formulas sin⁡(A+B)=sin⁡Acos⁡B+sin⁡Bcos⁡A\sin \left( A+B \right)=\sin A\cos B+\sin B\cos A.
So, after comparing the above expression sin⁡33∘cos⁡27∘+cos⁡33∘sin⁡27∘\sin 33{}^\circ \cos 27{}^\circ +\cos 33{}^\circ \sin 27{}^\circ with sin⁡Acos⁡B+sin⁡Bcos⁡A\sin A\cos B+\sin B\cos A we will get:
A=33∘A=33{}^\circ and B=27∘B=27{}^\circ .
So, we will put the value of A and B in the formula sin⁡(A+B)=sin⁡Acos⁡B+sin⁡Bcos⁡A\sin \left( A+B \right)=\sin A\cos B+\sin B\cos A, then we will get:
⇒sin⁡(33∘+27∘)=sin⁡33∘cos⁡27∘+sin⁡33∘cos⁡27∘\Rightarrow \sin \left( 33{}^\circ +27{}^\circ \right)=\sin 33{}^\circ \cos 27{}^\circ +\sin 33{}^\circ \cos 27{}^\circ
⇒sin⁡(60∘)=sin⁡33∘cos⁡27∘+sin⁡33∘cos⁡27∘\Rightarrow \sin \left( 60{}^\circ \right)=\sin 33{}^\circ \cos 27{}^\circ +\sin 33{}^\circ \cos 27{}^\circ
Now, from trigonometric table we know that sin⁡60∘=32\sin 60{}^\circ =\dfrac{\sqrt{3}}{2} , so we will put the value 32\dfrac{\sqrt{3}}{2} in place of sin⁡60∘\sin 60{}^\circ , then we will get:
⇒sin⁡33∘cos⁡27∘+sin⁡33∘cos⁡27∘=32\Rightarrow \sin 33{}^\circ \cos 27{}^\circ +\sin 33{}^\circ \cos 27{}^\circ =\dfrac{\sqrt{3}}{2}
So, we can say that exact value of sin⁡33∘cos⁡27∘+cos⁡33∘sin⁡27∘\sin 33{}^\circ \cos 27{}^\circ +\cos 33{}^\circ \sin 27{}^\circ is 32\dfrac{\sqrt{3}}{2}.
This is our required solution.

Note: Students are required to note that when we are given sec⁡θ\sec \theta , cosec⁡θ\operatorname{cosec}\theta , tan⁡θ\tan \theta , and cot⁡θ\cot \theta in the trigonometric expression then we always change them into sin⁡θ\sin \theta and cos⁡θ\cos \theta because we know half angle formula, double angle formula for sine and cosine function only so we by changing trigonometric function other than sine and cosine into sine and cosine make our take easier.