Questions & Answers

Find the value of x, if \[\tan 3x=\sin 45.\cos 45+\sin 30\].

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Hint: Form the table of trigonometric values with angles such as \[{{0}^{\circ }},{{30}^{\circ }},{{45}^{\circ }},{{60}^{\circ }},{{90}^{\circ }}\]and with trigonometric function sine, cosine, tangent. Find the value of trigonometric functions from the question and substitute to find the required answer.

Complete step-by-step answer:
Given, \[\tan 3x=\sin 45.\cos 45+\sin 30\].
We can find the values of RHS using basic trigonometric formulas.
We can find it by creating a trigonometric table with the required angles such as \[{{0}^{\circ }},{{30}^{\circ }},{{45}^{\circ }},{{60}^{\circ }}\]and \[{{90}^{\circ }}\]. With 6 trigonometric functions such as sine, cosine, tangent, cosecant, secant and cotangent.
We can draw the table.

Now considering the RHS = \[\sin 45.\cos 45+\sin 30\].
From the above table we can find the values of sin45, cos45 and sin30.
Value of \[\sin 45=\dfrac{1}{\sqrt{2}}\].
Value of \[\cos 45=\dfrac{1}{\sqrt{2}}\].
Value of \[\sin 30=\dfrac{1}{2}\].
Substituting the values in RHS we get,
  & RHS=\sin 45.\cos 45+\sin 30 \\
 & =\dfrac{1}{\sqrt{2}}\times \dfrac{1}{\sqrt{2}}+\dfrac{1}{2}=\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{2}{2}=1 \\
Given, \[LHS=\tan 3x\], putting the value of RHS, we get,
  & \tan 3x=1 \\
 & \Rightarrow 3x{{\tan }^{-1}}\left( 1 \right) \\
From the table, \[\tan 45=1\].
Taking the \[{{\tan }^{-1}}\left( 1 \right)\]we get \[{{45}^{\circ }}\].
  & \therefore 3x=45 \\
 & \Rightarrow x=\dfrac{45}{3}={{15}^{\circ }} \\
Therefore, we get the values of x as \[{{15}^{\circ }}\].
We got \[{{\tan }^{-1}}\left( 1 \right)={{45}^{\circ }}\].
We know \[{{\tan }^{-1}}\theta =\cot \theta \]i.e. \[\dfrac{1}{\tan \theta }=\cot \theta \].
From the table of the trigonometric functions. Or try to remember the first 3 functions \[\sin \theta ,\cos \theta \] and \[\tan \theta \]. The other 3 functions can be found taking the reverse of \[\sin \theta \] (to get \[\cos ec\theta \]), \[\cos \theta \] (to get \[\sec \theta \]) and \[\tan \theta \] (to get \[\cot \theta \]).
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