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Question: How do you evaluate \(\cos 150{}^\circ\)?...

How do you evaluate cos⁡150∘\cos 150{}^\circ?

Explanation

Solution

In this problem we have to evaluate cos⁡150∘\cos 150{}^\circ. For this type of problem, we will write the given angle as the sum or difference of two angles for which we have the values of the trigonometric ratios. Now according to the trigonometric ratio, we will use the appropriate formula and simplify the equation to get the result. For this problem, we will write the given angle 150∘150{}^\circ as the sum of the 90∘90{}^\circ, 60∘60{}^\circ. Now we will use the trigonometric formula cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B. In this formula, we will apply all the trigonometric values we have and simplify them to get the required result.

Formula use:
1. we have the trigonometric 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:
Given that, cos⁡150∘\cos 150{}^\circ.
Consider the given angle 150∘150{}^\circ . We can write the given angle 150∘150{}^\circ as sum of the 90∘90{}^\circ and 60∘60{}^\circ . Mathematically we can write
150∘=90∘+60∘150{}^\circ =90{}^\circ +60{}^\circ
Applying the trigonometric function cos⁡\cos on both sides of the above equation, then we will get
⇒cos⁡150∘=cos⁡(60∘+90∘)\Rightarrow \cos 150{}^\circ =\cos \left( 60{}^\circ +90{}^\circ \right)
Using the trigonometric formula cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B in the above equation, then we will get
⇒cos⁡(60∘+90∘)=cos⁡60∘.cos⁡90∘−sin⁡60∘sin⁡90∘\Rightarrow \cos \left( 60{}^\circ +90{}^\circ \right)=\cos 60{}^\circ .\cos 90{}^\circ -\sin 60{}^\circ \sin 90{}^\circ
To solve the above equation, we need to have the values of two trigonometric ratios which are sin⁡\sin and cos⁡\cos.
Considering the cos⁡\cos terms in the above equation.
We have cos⁡60∘\cos 60{}^\circ , cos⁡90∘\cos 90{}^\circ.
We know that cos⁡60∘=12\cos 60{}^\circ =\dfrac{1}{2} and also, we know that cos⁡90∘=0\cos 90{}^\circ =0.
Now substituting these values in the above equation, then we will get
⇒cos⁡(60∘+90∘)=12.0−sin⁡60∘sin⁡90∘\Rightarrow \cos \left( 60{}^\circ +90{}^\circ \right)=\dfrac{1}{2}.0-\sin 60{}^\circ \sin 90{}^\circ
We know that the product of anything with the zero will be zero, then we will get
⇒cos⁡(60∘+90∘)=0−sin⁡60∘sin⁡90∘ ⇒cos⁡(60∘+90∘)=−sin⁡60∘sin⁡90∘ \begin{aligned} & \Rightarrow \cos \left( 60{}^\circ +90{}^\circ \right)=0-\sin 60{}^\circ \sin 90{}^\circ \\\ & \Rightarrow \cos \left( 60{}^\circ +90{}^\circ \right)=-\sin 60{}^\circ \sin 90{}^\circ \\\ \end{aligned}
Considering the sin⁡\sin terms in the above equation.
We have sin⁡60∘\sin 60{}^\circ , sin⁡90∘\sin 90{}^\circ.
We know that sin⁡60∘=32\sin 60{}^\circ =\dfrac{\sqrt{3}}{2} and also, we know that sin⁡90∘=1\sin 90{}^\circ =1
Substituting these values in the above equation, then we will get
⇒cos⁡(60∘+90∘)=−32×1\Rightarrow \cos \left( 60{}^\circ +90{}^\circ \right)=-\dfrac{\sqrt{3}}{2}\times 1

Simplifying the above equation, we have the value of cos⁡150∘\cos 150{}^\circ as −32-\dfrac{\sqrt{3}}{2}.

Note:
We can find the value of the above expression in another method. We can use property trigonometric table, unit circle and property of complementary arc. That is
cos⁡150∘=cos⁡(90∘+60∘)\cos 150{}^\circ =\cos \left( 90{}^\circ +60{}^\circ \right)
We know that cos⁡(90∘+θ)=−sin⁡θ\cos \left( 90{}^\circ +\theta \right)=-\sin \theta
⇒cos⁡150∘=−sin⁡60∘\Rightarrow \cos 150{}^\circ =-\sin 60{}^\circ
We know that sin⁡60∘=32\sin 60{}^\circ =\dfrac{\sqrt{3}}{2} then we will get final result that is
⇒cos⁡150∘=−32\Rightarrow \cos 150{}^\circ =-\dfrac{\sqrt{3}}{2}.