
How do you find the exact values of \[\tan {{150}^{\circ }}\] using half angle identity?
Answer
523.5k+ views
Hint: In this problem, we have to find the exact value of \[\tan {{150}^{\circ }}\] using the half angle formula. We know that basically half angle formulas are derived from the sum of sine, cosine and tangent identities. By using the half angle formula \[\tan 2t=\dfrac{2\tan t}{1-{{\tan }^{2}}t}\], we can find the given tangent value. In this formula we have to find the \[\tan 2t\] , and substitute it in the identity to get the exact value.
Complete step by step solution:
We know that the given angle is, \[\tan {{150}^{\circ }}\]
We also know that one of the half angle identities is
\[\tan 2t=\dfrac{2\tan t}{1-{{\tan }^{2}}t}\]
We know that the given degree is 150, where we can assume it as 150 degrees.
\[\begin{align}
& \Rightarrow \tan 2t=\tan {{300}^{\circ }}=\tan \left( -60+360 \right) \\
& \Rightarrow \tan 2t=\tan \left( -60 \right)=-\sqrt{3} \\
\end{align}\]
Now we can apply the half angle formula for \[\tan {{15}^{\circ }}\] as we have \[\tan 2t=-\sqrt{3}\] .
\[\Rightarrow \dfrac{2\tan t}{1-{{\tan }^{2}}t}=-\sqrt{3}\]
We can now cross multiply the above step, we get
\[\begin{align}
& \Rightarrow 2\tan t=-\sqrt{3}+\sqrt{3}{{\tan }^{2}}t \\
& \Rightarrow \sqrt{3}{{\tan }^{2}}t-2\tan t-\sqrt{3}=0 \\
\end{align}\]
We can solve this quadratic equation, we get using the quadratic formula
Where a = \[\sqrt{3}\], b = -2, c = \[-\sqrt{3}\] , we get
\[\begin{align}
& \Rightarrow \tan t=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{2\pm \sqrt{4-4\left( -\sqrt{3} \right)\left( \sqrt{3} \right)}}{2\sqrt{3}} \\
& \Rightarrow \tan t=\dfrac{2\pm \sqrt{16}}{2\sqrt{3}}=\dfrac{2\pm 4}{2\sqrt{3}} \\
\end{align}\]
Here we have two roots, where we can ignore the positive term as \[\tan {{150}^{\circ }}\]is negative , we get
\[\Rightarrow \tan t=\tan {{150}^{\circ }}=-\dfrac{1}{\sqrt{3}}=-0.577\]
Therefore, the exact value of \[\tan {{150}^{\circ }}\] using half angle formula is -0.577.
Note: Students make mistakes while applying the \[\tan 2t\] values. We should know that to solve these types of problems, we have to know basic trigonometric identities, formulas, degree values and properties. We should also know that we have different methods to find the value using the half angle formula. We should also know some root values to be applied in these types of problems.
Complete step by step solution:
We know that the given angle is, \[\tan {{150}^{\circ }}\]
We also know that one of the half angle identities is
\[\tan 2t=\dfrac{2\tan t}{1-{{\tan }^{2}}t}\]
We know that the given degree is 150, where we can assume it as 150 degrees.
\[\begin{align}
& \Rightarrow \tan 2t=\tan {{300}^{\circ }}=\tan \left( -60+360 \right) \\
& \Rightarrow \tan 2t=\tan \left( -60 \right)=-\sqrt{3} \\
\end{align}\]
Now we can apply the half angle formula for \[\tan {{15}^{\circ }}\] as we have \[\tan 2t=-\sqrt{3}\] .
\[\Rightarrow \dfrac{2\tan t}{1-{{\tan }^{2}}t}=-\sqrt{3}\]
We can now cross multiply the above step, we get
\[\begin{align}
& \Rightarrow 2\tan t=-\sqrt{3}+\sqrt{3}{{\tan }^{2}}t \\
& \Rightarrow \sqrt{3}{{\tan }^{2}}t-2\tan t-\sqrt{3}=0 \\
\end{align}\]
We can solve this quadratic equation, we get using the quadratic formula
Where a = \[\sqrt{3}\], b = -2, c = \[-\sqrt{3}\] , we get
\[\begin{align}
& \Rightarrow \tan t=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{2\pm \sqrt{4-4\left( -\sqrt{3} \right)\left( \sqrt{3} \right)}}{2\sqrt{3}} \\
& \Rightarrow \tan t=\dfrac{2\pm \sqrt{16}}{2\sqrt{3}}=\dfrac{2\pm 4}{2\sqrt{3}} \\
\end{align}\]
Here we have two roots, where we can ignore the positive term as \[\tan {{150}^{\circ }}\]is negative , we get
\[\Rightarrow \tan t=\tan {{150}^{\circ }}=-\dfrac{1}{\sqrt{3}}=-0.577\]
Therefore, the exact value of \[\tan {{150}^{\circ }}\] using half angle formula is -0.577.
Note: Students make mistakes while applying the \[\tan 2t\] values. We should know that to solve these types of problems, we have to know basic trigonometric identities, formulas, degree values and properties. We should also know that we have different methods to find the value using the half angle formula. We should also know some root values to be applied in these types of problems.
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