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Question: The value of \(\cos 30^\circ \cos 60^\circ - \sin 30^\circ \sin 60^\circ = \)...

The value of cos⁡30∘cos⁡60∘−sin⁡30∘sin⁡60∘=\cos 30^\circ \cos 60^\circ - \sin 30^\circ \sin 60^\circ =

Explanation

Solution

Here, we are required to find the value of a given expression. We will use the formula of cos⁡(A+B)\cos \left( {A + B} \right) and substitute the given measure of angles in the formula. Then we will simplify it further to get the required value.

Formula Used:
We will use the formula cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B.

Complete step by step solution:
We can observe that the given expression is in the form cos⁡Acos⁡B−sin⁡Asin⁡B\cos A\cos B - \sin A\sin B.
Now comparing the given expression with cos⁡Acos⁡B−sin⁡Asin⁡B\cos A\cos B - \sin A\sin B, we get
A=60∘A = 60^\circ and B=30∘B = 30^\circ
Substituting A=60∘A = 60^\circ and B=30∘B = 30^\circ in the formula cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B, we get
cos⁡30∘cos⁡60∘−sin⁡30∘sin⁡60∘=cos⁡(60∘+30∘)\cos 30^\circ \cos 60^\circ - \sin 30^\circ \sin 60^\circ = \cos \left( {60^\circ + 30^\circ } \right)
⇒cos⁡30∘cos⁡60∘−sin⁡30∘sin⁡60∘=cos⁡(90∘)\Rightarrow \cos 30^\circ \cos 60^\circ - \sin 30^\circ \sin 60^\circ = \cos \left( {90^\circ } \right)
Now we know that cos⁡90∘=0\cos 90^\circ = 0, therefore, we get
⇒cos⁡30∘cos⁡60∘−sin⁡30∘sin⁡60∘=0\Rightarrow \cos 30^\circ \cos 60^\circ - \sin 30^\circ \sin 60^\circ = 0

Therefore, the value of cos⁡30∘cos⁡60∘−sin⁡30∘sin⁡60∘=0\cos 30^\circ \cos 60^\circ - \sin 30^\circ \sin 60^\circ = 0.

Additional information:
Trigonometry is a branch of mathematics which helps us to study the relationship between the sides and the angles of a triangle. In practical life, trigonometry is used by cartographers (to make maps). It is also used by the aviation and naval industries. In fact, trigonometry is even used by Astronomers to find the distance between two stars.
Hence, it has an important role to play in everyday life.

Note:
The most common mistake which we can do while solving this question is that in the formula we can take cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B as cos⁡(A−B)\cos \left( {A - B} \right). This is because, in the RHS, we have a ‘minus sign’ and it will confuse us to use the ‘minus sign’ in the LHS as well. Hence, it will make our answer wrong.
Also, we can solve this question without using the formulas by just substituting the values from the trigonometric table show below:

| 0∘0^\circ | 30∘30^\circ | 45∘45^\circ | 60∘60^\circ | 90∘90^\circ
---|---|---|---|---|---
sin⁡\sin | 0| 12\dfrac{1}{2}| 12\dfrac{1}{{\sqrt 2 }}| 32\dfrac{{\sqrt 3 }}{2}| 1
cos⁡\cos | 1| 32\dfrac{{\sqrt 3 }}{2}| 12\dfrac{1}{{\sqrt 2 }}| 12\dfrac{1}{2}| 0
tan⁡\tan | 0| 13\dfrac{1}{{\sqrt 3 }}| 1| 3\sqrt 3 | Not Defined

Hence, the value of cos⁡30∘cos⁡60∘−sin⁡30∘sin⁡60∘=(32)⋅(12)−(12)⋅(32)=0\cos 30^\circ \cos 60^\circ - \sin 30^\circ \sin 60^\circ = \left( {\dfrac{{\sqrt 3 }}{2}} \right) \cdot \left( {\dfrac{1}{2}} \right) - \left( {\dfrac{1}{2}} \right) \cdot \left( {\dfrac{{\sqrt 3 }}{2}} \right) = 0
Therefore, the value of cos⁡30∘cos⁡60∘−sin⁡30∘sin⁡60∘=0\cos 30^\circ \cos 60^\circ - \sin 30^\circ \sin 60^\circ = 0.