# 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

•

Total views: 307.2k

•

Views today: 4.84k

Answer

Verified

307.2k+ views

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.

Complete step-by-step answer:

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:

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.

Complete step-by-step answer:

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.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE