
Find \[\tan 22.5\] using a half- angle formula.
Answer
501.3k+ views
Hint: In this problem, we have to find the value of \[\tan 22.5\] using a half -angle formula. We will use the half angle formula for tangent as follows.
\[\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\]
We will put \[\theta = 22.5\] in a half angle formula and then solve by cross multiplying to get the desired value. We will also use the value of \[\tan 45 = 1\] .
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, \[\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\]
If we put \[A = B\] then the formula becomes
\[\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}\]
Considering the given question, we have to find the value of \[\tan 22.5\] .
From half angle formula we have,
\[\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\]
Let, \[\theta = 22.5\] Then \[2\theta = 45\] .
Putting \[\theta = 22.5\] in above half angle formula we get
\[\tan 45 = \dfrac{{2\tan 22.5}}{{1 - {{\tan }^2}22.5}}\]
We know that \[\tan 45 = 1\]
Hence we have,
\[\dfrac{{2\tan 22.5}}{{1 - {{\tan }^2}22.5}} = 1\]
On cross multiplication we have,
\[2\tan 22.5 = 1 - {\tan ^2}22.5\]
Adding \[{\tan ^2}22.5 - 1\] to both sides, we have,
\[{\tan ^2}22.5 + 2\tan 22.5 - 1 = 0\]
Let , \[x = \tan 22.5\] then we have,
\[{x^2} + 2x - 1 = 0\]
This is a quadratic equation. We know that the solution of quadratic equation \[a{x^2} + bx + c = 0\] is given by \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .
Here, \[a = 1\] , \[b = 2\] and \[c = - 1\] .
Hence , \[x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4 \times 1 \times ( - 1)} }}{{2 \times 1}} = \dfrac{{ - 2 \pm 2\sqrt 2 }}{2}\]
Hence , \[x = - 1 \pm \sqrt 2 \] .
Therefore, \[\tan 22.5 = - 1 + \sqrt 2 \] or \[\tan 22.5 = - 1 - \sqrt 2 \]
Since , \[\theta = 22.5\] lies in the first quadrant .
Therefore the value of \[\tan 22.5\] is positive.
Hence, \[\tan 22.5 = - 1 + \sqrt 2 \]
Hence the value of \[\tan 22.5\] is \[\sqrt 2 - 1\]
So, the correct answer is “ \[\sqrt 2 - 1\] ”.
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
\[\sin 2\theta = 2\sin \theta \cos \theta \]
\[\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta \]
\[\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\]
We will put \[\theta = 22.5\] in a half angle formula and then solve by cross multiplying to get the desired value. We will also use the value of \[\tan 45 = 1\] .
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, \[\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\]
If we put \[A = B\] then the formula becomes
\[\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}\]
Considering the given question, we have to find the value of \[\tan 22.5\] .
From half angle formula we have,
\[\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\]
Let, \[\theta = 22.5\] Then \[2\theta = 45\] .
Putting \[\theta = 22.5\] in above half angle formula we get
\[\tan 45 = \dfrac{{2\tan 22.5}}{{1 - {{\tan }^2}22.5}}\]
We know that \[\tan 45 = 1\]
Hence we have,
\[\dfrac{{2\tan 22.5}}{{1 - {{\tan }^2}22.5}} = 1\]
On cross multiplication we have,
\[2\tan 22.5 = 1 - {\tan ^2}22.5\]
Adding \[{\tan ^2}22.5 - 1\] to both sides, we have,
\[{\tan ^2}22.5 + 2\tan 22.5 - 1 = 0\]
Let , \[x = \tan 22.5\] then we have,
\[{x^2} + 2x - 1 = 0\]
This is a quadratic equation. We know that the solution of quadratic equation \[a{x^2} + bx + c = 0\] is given by \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .
Here, \[a = 1\] , \[b = 2\] and \[c = - 1\] .
Hence , \[x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4 \times 1 \times ( - 1)} }}{{2 \times 1}} = \dfrac{{ - 2 \pm 2\sqrt 2 }}{2}\]
Hence , \[x = - 1 \pm \sqrt 2 \] .
Therefore, \[\tan 22.5 = - 1 + \sqrt 2 \] or \[\tan 22.5 = - 1 - \sqrt 2 \]
Since , \[\theta = 22.5\] lies in the first quadrant .
Therefore the value of \[\tan 22.5\] is positive.
Hence, \[\tan 22.5 = - 1 + \sqrt 2 \]
Hence the value of \[\tan 22.5\] is \[\sqrt 2 - 1\]
So, the correct answer is “ \[\sqrt 2 - 1\] ”.
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
\[\sin 2\theta = 2\sin \theta \cos \theta \]
\[\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta \]
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