Solveeit Logo

Question

Question: Solve the following: \(\tan (1/2{{\sin }^{-1}}3/4)\)...

Solve the following: tan(1/2sin13/4)\tan (1/2{{\sin }^{-1}}3/4)

Explanation

Solution

Revise all the formulas of trigonometry and all the properties of inverse trigonometric functions. The inverse trigonometric functions are the inverse functions of the trigonometric functions for example inverse of sine, cosine, tangent etc.

Complete step by step solution:
We have to solve tan(1/2sin13/4)\tan (1/2{{\sin }^{-1}}3/4) for that let us assume that
1/2sin13/4=θ1/2{{\sin }^{-1}}3/4=\theta
Now by cross multiplication
1/2sin13/4=θ1/2{{\sin }^{-1}}3/4=\theta becomes
sin13/4=2θ{{\sin }^{-1}}3/4=2\theta ---- (1)
Now by multiplying sin on both sides of the equation(1)
sin(sin13/4)=sin2θ\sin ({{\sin }^{-1}}3/4)=\sin 2\theta
sin2θ=3/4\sin 2\theta =3/4 ---- (2)
Now by using the formula sin2θ=2tanθ/(1+tan2θ)\sin 2\theta =2\tan \theta /(1+{{\tan }^{2}}\theta ) on equation (2)
2tanθ/(1+tan2θ)=3/42\tan \theta /(1+{{\tan }^{2}}\theta )=3/4
Now by cross multiplication
2tanθ/(1+tan2θ)=3/42\tan \theta /(1+{{\tan }^{2}}\theta )=3/4 becomes
4(2tanθ)=3(1+tan2θ)4(2\tan \theta )=3(1+{{\tan }^{2}}\theta ) ----- (3)
By solving the brackets of equation (3)
8tanθ=3+3tan2θ8\tan \theta =3+3{{\tan }^{2}}\theta
3tan2θ8tanθ+3=03{{\tan }^{2}}\theta -8\tan \theta +3=0 ------- (4)
To find the roots of a quadratic equation we use the formula
x=b±b24ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
So by using the above formula on equation (4)
tanθ=(8)±64(4×3×3/2×3\tan \theta =-(-8)\pm \sqrt{64-(4 \times 3 \times 3}/2 \times 3
So, tanθ=4±7/3\tan \theta =4\pm \sqrt{7}/3 ----- (5)
Now by taking tan inverse on both sides of equation (5)
θ=tan1[4±7/3]\theta ={{\tan }^{-1}}\left[ 4\pm \sqrt{7}/3 \right]
tanθ=4±7/3\tan \theta =4\pm \sqrt{7}/3
As θ=1/2sin13/4\theta =1/2{{\sin }^{-1}}3/4
So, tanθ=4±7/3\tan \theta =4\pm \sqrt{7}/3
Since ,
π/2sin13/4π/2 π/41/2sin13/4π/4 \begin{aligned} & -\pi /2\le {{\sin }^{-1}}3/4\le \pi /2 \\\ & -\pi /4\le 1/2{{\sin }^{-1}}3/4\le \pi /4 \\\ \end{aligned}
Therefore, tan(π/4)tan1/2(sin13/4)tanπ/4\tan (-\pi /4)\le \tan 1/2({{\sin }^{-1}}3/4)\le \tan \pi /4
1tan(1/2sin13/4)1-1\le \tan (1/2{{\sin }^{-1}}3/4)\le 1
Since, 4+7/3>14+\sqrt{7}/3>1 so it is ignored
Therefore, tan(1/2sin13/4)=47/3\tan (1/2{{\sin }^{-1}}3/4)=4-\sqrt{7}/3.

Note:
There is a restriction on sinθ\sin \theta i.e. π/2sin1θπ/2-\pi /2\le {{\sin }^{-1}}\theta \le \pi /2. So all the values which are greater than one should be ignored. Always use the correct trigonometric formula to solve a particular equation as using the wrong formula leads towards the wrong answer.