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# Find the value of x, if x satisfies following trigonometric equation:${{\cos }^{2}}45-{{\cos }^{2}}30=x.\cos 45.\sin 45$,

Last updated date: 29th Mar 2023
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Hint: Here, first of all, substitute the values of cos 45, cos 30 and sin 45 in the given equation and then solve the equation to get the required value of x. So, use this method to solve the question.

We are given that ${{\cos }^{2}}45-{{\cos }^{2}}30=x.\cos 45.\sin 45$. Here, we have to find the value of x.
We know that we already have the value of trigonometric ratios like $\sin \theta ,\cos \theta$ etc. of some basic angles like ${{30}^{o}}{{45}^{o}},{{0}^{o}},{{90}^{o}}$ etc.
We can get them by referring to the table for basic trigonometric ratios in which we can find the values of trigonometric ratios that are $\sin \theta ,\cos \theta ,\tan \theta ,\cot \theta, \operatorname{cosec}\theta$ and $\cot \theta$ at different angles that are ${{0}^{o}},{{30}^{o}},{{60}^{o}},{{45}^{o}}$ and ${{90}^{o}}$.
Our table is as follows:

 Angles / Trigonometric Ratio $\sin \theta$ $\cos \theta$ $\tan \theta$ $\cot \theta$ $sec\theta$ $\operatorname{cosec}\theta$ ${{0}^{o}}$ 0 1 0 NA 1 NA ${{30}^{o}}$ $\dfrac{1}{2}$ $\dfrac{\sqrt{3}}{2}$ $\dfrac{1}{\sqrt{3}}$ $\sqrt{3}$ $\dfrac{2}{\sqrt{3}}$ 2 ${{45}^{o}}$ $\dfrac{1}{\sqrt{2}}$ $\dfrac{1}{\sqrt{2}}$ 1 1 $\sqrt{2}$ $\sqrt{2}$ ${{60}^{o}}$ $\dfrac{\sqrt{3}}{2}$ $\dfrac{1}{2}$ $\sqrt{3}$ $\dfrac{1}{\sqrt{3}}$ 2 $\dfrac{2}{\sqrt{3}}$ ${{90}^{o}}$ 1 0 NA 0 NA 1

Now, let us consider the equation given in the question
${{\cos }^{2}}45-{{\cos }^{2}}30=x.\cos 45.\sin 45....\left( i \right)$
Now, from the above table, we can find the values of $\cos {{45}^{o}},\cos {{30}^{o}}$ and $\sin {{45}^{o}}$.
We get,
Value of $\sin {{45}^{o}}=\dfrac{1}{\sqrt{2}}$
Value of $\cos {{45}^{o}}=\dfrac{1}{\sqrt{2}}$
Value of $\cos {{30}^{o}}=\dfrac{\sqrt{3}}{2}$
By substituting the values, $\sin {{45}^{o}}.\cos {{45}^{o}}$ and $\cos {{30}^{o}}$ in equation (i), we get,
${{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}-{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}=x\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{1}{\sqrt{2}} \right)$
By simplifying the above equation, we get,
$\dfrac{1}{2}-\dfrac{3}{2}=x\left( \dfrac{1}{2} \right)$
Or, $\dfrac{\left( 1-3 \right)}{2}=\dfrac{x}{2}$
By multiplying 2 on both sides, we get,
$\Rightarrow 2\left( \dfrac{-2}{2} \right)=2\left( \dfrac{x}{2} \right)$
$\Rightarrow -2=x$
Or, $x=-2$
Hence, we get the value of x = -2
Note: Students are advised to remember the values of at least the first 2 trigonometric ratios that are $\sin \theta$ and $\cos \theta$ at different angles. Other ratios can be found with these as we can get $\tan \theta =\dfrac{\sin \theta }{\cos \theta },\cot \theta =\dfrac{\cos \theta }{\sin \theta },\operatorname{cosec}\theta =\dfrac{1}{\sin \theta },\sec \theta =\dfrac{1}{\cos \theta }$. In this question, students can cross-check their answer by substituting the value of x in the given expression and verifying LHS=RHS.