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Question

Question: Find \[\tan 22.5\] using a half- angle formula....

Find tan22.5\tan 22.5 using a half- angle formula.

Explanation

Solution

Hint : In this problem, we have to find the value of tan22.5\tan 22.5 using a half -angle formula. We will use the half angle formula for tangent as follows.
tan2θ=2tanθ1tan2θ\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}
We will put θ=22.5\theta = 22.5 in a half angle formula and then solve by cross multiplying to get the desired value. We will also use the value of tan45=1\tan 45 = 1 .

Complete step by step solution:
This question is based on application of trigonometric formulas. Trigonometric formula is based on the relationship between T-ratio of angles, identify sides etc.
Half angle formula is based on the formula for the sum of two angles.
For example, tan(A+B)=tanA+tanB1tanAtanB\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}
If we put A=BA = B then the formula becomes
tan2A=2tanA1tan2A\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}
Considering the given question, we have to find the value of tan22.5\tan 22.5 .
From half angle formula we have,
tan2θ=2tanθ1tan2θ\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}
Let, θ=22.5\theta = 22.5 Then 2θ=452\theta = 45 .
Putting θ=22.5\theta = 22.5 in above half angle formula we get
tan45=2tan22.51tan222.5\tan 45 = \dfrac{{2\tan 22.5}}{{1 - {{\tan }^2}22.5}}
We know that tan45=1\tan 45 = 1
Hence we have,
2tan22.51tan222.5=1\dfrac{{2\tan 22.5}}{{1 - {{\tan }^2}22.5}} = 1
On cross multiplication we have,
2tan22.5=1tan222.52\tan 22.5 = 1 - {\tan ^2}22.5
Adding tan222.51{\tan ^2}22.5 - 1 to both sides, we have,
tan222.5+2tan22.51=0{\tan ^2}22.5 + 2\tan 22.5 - 1 = 0
Let , x=tan22.5x = \tan 22.5 then we have,
x2+2x1=0{x^2} + 2x - 1 = 0
This is a quadratic equation. We know that the solution of quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 is given by x=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} .
Here, a=1a = 1 , b=2b = 2 and c=1c = - 1 .
Hence , x=2±224×1×(1)2×1=2±222x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4 \times 1 \times ( - 1)} }}{{2 \times 1}} = \dfrac{{ - 2 \pm 2\sqrt 2 }}{2}
Hence , x=1±2x = - 1 \pm \sqrt 2 .
Therefore, tan22.5=1+2\tan 22.5 = - 1 + \sqrt 2 or tan22.5=12\tan 22.5 = - 1 - \sqrt 2
Since , θ=22.5\theta = 22.5 lies in the first quadrant .
Therefore the value of tan22.5\tan 22.5 is positive.
Hence, tan22.5=1+2\tan 22.5 = - 1 + \sqrt 2
Hence the value of tan22.5\tan 22.5 is 21\sqrt 2 - 1
So, the correct answer is “ 21\sqrt 2 - 1 ”.

Note : Values of all T-ratios are positive in the first quadrant. While In second quadrant, only the value of sine is positive. In the third quadrant, only tangent is positive and in the fourth quadrant, only cosine is positive.
Quadratic equations can also be solved by splitting the middle term such that sum of terms is middle term and product is constant term.
Some important trigonometry half angle formula are
sin2θ=2sinθcosθ\sin 2\theta = 2\sin \theta \cos \theta
cos2θ=cos2θsin2θ\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta