Answer
438.6k+ views
Hint: The values of trigonometric ratios of standard angles are known to us from the trigonometric tables. Using the trigonometric table substitute the values of $\cos 90{}^\circ =0\text{ and cos30}{}^\circ \text{=}\dfrac{\sqrt{3}}{2}$ in the expressions that you get after substituting $\theta =30{}^\circ $ in the RHS and the LHS of the equation given in the question. If the values that you get from the RHS and the LHS are equal, you can say that you have verified the equation.
Complete step by step solution:
Before moving to the solution, let us discuss the nature of sine and cosine function, which we would be using in the solution. We can better understand this using the graph of sine and cosine.
First, let us start with the graph of sinx.
Next, let us see the graph of cosx.
Looking at both the graphs, and using the relations between the different trigonometric ratios, we get
Let us start the solution to the above question by simplifying the left-hand side of the equation $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta $ by putting the value $\theta =30{}^\circ $ .
$\cos 3\theta $
$=\cos \left( 3\times 30{}^\circ \right)$
$=\cos 90{}^\circ $
And we know that the value of $\cos 90{}^\circ $ is equal to 0. So, we get
$=\cos 90{}^\circ =0$
So, the left-hand side of the equation $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta $ is equal to $0$ .
Now let us simplify the right-hand side of the equation by putting the values $\theta =30{}^\circ $ .
$4{{\cos }^{3}}\theta -3\cos \theta $
$=4{{\cos }^{3}}30{}^\circ -3\cos 30{}^\circ $
Now we know that $\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}$ .
$=4{{\cos }^{3}}30{}^\circ -3\cos 30{}^\circ $
$=4\times {{\left( \dfrac{\sqrt{3}}{2} \right)}^{3}}-3\times \dfrac{\sqrt{3}}{2}$
$=\dfrac{\sqrt{3}}{2}\left( 4\times \dfrac{3}{4}-3 \right)=0$
Therefore, we can say that the left-hand side of the equation $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta $ is equal to the right-hand side. So, we have verified the above equation.
Note: It is generally seen that students are very confused in the values of $\sin 30{}^\circ ,\sin 60{}^\circ ,\cos 30{}^\circ \text{ and cos60}{}^\circ $ and very frequently make mistakes in their values. So, it is very important that you thoroughly learn the trigonometric table and properties related to complementary angles and specially the values of $\sin 30{}^\circ ,\sin 60{}^\circ ,\cos 30{}^\circ \text{ and cos60}{}^\circ $ .
Complete step by step solution:
Before moving to the solution, let us discuss the nature of sine and cosine function, which we would be using in the solution. We can better understand this using the graph of sine and cosine.
First, let us start with the graph of sinx.
![seo images](https://www.vedantu.com/question-sets/b7fc54bf-126a-41ab-9459-1ad21168e0608849037646545734903.png)
Next, let us see the graph of cosx.
![seo images](https://www.vedantu.com/question-sets/6b19b30d-1c51-42dd-9014-d25c49d4b8a24326934078757579188.png)
Looking at both the graphs, and using the relations between the different trigonometric ratios, we get
![seo images](https://www.vedantu.com/question-sets/bed06615-c0a9-45f1-9790-1d4c655ec9291254174284014269005.png)
Let us start the solution to the above question by simplifying the left-hand side of the equation $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta $ by putting the value $\theta =30{}^\circ $ .
$\cos 3\theta $
$=\cos \left( 3\times 30{}^\circ \right)$
$=\cos 90{}^\circ $
And we know that the value of $\cos 90{}^\circ $ is equal to 0. So, we get
$=\cos 90{}^\circ =0$
So, the left-hand side of the equation $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta $ is equal to $0$ .
Now let us simplify the right-hand side of the equation by putting the values $\theta =30{}^\circ $ .
$4{{\cos }^{3}}\theta -3\cos \theta $
$=4{{\cos }^{3}}30{}^\circ -3\cos 30{}^\circ $
Now we know that $\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}$ .
$=4{{\cos }^{3}}30{}^\circ -3\cos 30{}^\circ $
$=4\times {{\left( \dfrac{\sqrt{3}}{2} \right)}^{3}}-3\times \dfrac{\sqrt{3}}{2}$
$=\dfrac{\sqrt{3}}{2}\left( 4\times \dfrac{3}{4}-3 \right)=0$
Therefore, we can say that the left-hand side of the equation $\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta $ is equal to the right-hand side. So, we have verified the above equation.
Note: It is generally seen that students are very confused in the values of $\sin 30{}^\circ ,\sin 60{}^\circ ,\cos 30{}^\circ \text{ and cos60}{}^\circ $ and very frequently make mistakes in their values. So, it is very important that you thoroughly learn the trigonometric table and properties related to complementary angles and specially the values of $\sin 30{}^\circ ,\sin 60{}^\circ ,\cos 30{}^\circ \text{ and cos60}{}^\circ $ .
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