How do you find the exact values of \[\tan {{15}^{\circ }}\] using half angle formulas?
Answer
578.1k+ views
Hint: In this problem, we have to find the exact value of \[\tan {{15}^{\circ }}\] 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 \[\tan \dfrac{\theta }{2}=\dfrac{1-\cos \theta }{\sin \theta }\], we can find the given tangent value. In this formula we have to find the theta value, and substitute the exact sine and cosine degree values, to get the exact value.
Complete step by step answer:
We know that the given angle is,\[\tan {{15}^{\circ }}\]
We also know that one of the half angle formulas is
\[\tan \dfrac{\theta }{2}=\dfrac{1-\cos \theta }{\sin \theta }\]
We know that the given degree is 15, where
\[\begin{align}
& \Rightarrow \dfrac{\theta }{2}={{15}^{\circ }} \\
& \Rightarrow \theta ={{30}^{\circ }} \\
\end{align}\]
Now we can apply the half angle formula for \[\tan {{15}^{\circ }}\] as we have \[\theta ={{30}^{\circ }}\] .
\[\Rightarrow \tan {{15}^{\circ }}=\dfrac{1-\cos {{30}^{\circ }}}{\sin {{30}^{\circ }}}\] …… (1)
We know that the value of \[\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}\] and \[\sin {{30}^{\circ }}=\dfrac{1}{2}\]
Applying the above values in expression (1), we get
\[\begin{align}
& \Rightarrow \tan {{15}^{\circ }}=\dfrac{1-\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} \\
& \Rightarrow \tan {{15}^{\circ }}=\dfrac{\dfrac{2-\sqrt{3}}{2}}{\dfrac{1}{2}} \\
& \Rightarrow \tan {{15}^{\circ }}=2-\sqrt{3} \\
\end{align}\]
We know that the value of \[\sqrt{3}=1.732\], applying this value in the above step to get the exact value, we get
\[\Rightarrow \tan {{15}^{\circ }}=2-1.732=0.267\]
Therefore, the exact value of \[\tan {{15}^{\circ }}\] using half angle formula is 0.267.
Note: Students make mistakes while applying the sine and cosine 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 answer:
We know that the given angle is,\[\tan {{15}^{\circ }}\]
We also know that one of the half angle formulas is
\[\tan \dfrac{\theta }{2}=\dfrac{1-\cos \theta }{\sin \theta }\]
We know that the given degree is 15, where
\[\begin{align}
& \Rightarrow \dfrac{\theta }{2}={{15}^{\circ }} \\
& \Rightarrow \theta ={{30}^{\circ }} \\
\end{align}\]
Now we can apply the half angle formula for \[\tan {{15}^{\circ }}\] as we have \[\theta ={{30}^{\circ }}\] .
\[\Rightarrow \tan {{15}^{\circ }}=\dfrac{1-\cos {{30}^{\circ }}}{\sin {{30}^{\circ }}}\] …… (1)
We know that the value of \[\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}\] and \[\sin {{30}^{\circ }}=\dfrac{1}{2}\]
Applying the above values in expression (1), we get
\[\begin{align}
& \Rightarrow \tan {{15}^{\circ }}=\dfrac{1-\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} \\
& \Rightarrow \tan {{15}^{\circ }}=\dfrac{\dfrac{2-\sqrt{3}}{2}}{\dfrac{1}{2}} \\
& \Rightarrow \tan {{15}^{\circ }}=2-\sqrt{3} \\
\end{align}\]
We know that the value of \[\sqrt{3}=1.732\], applying this value in the above step to get the exact value, we get
\[\Rightarrow \tan {{15}^{\circ }}=2-1.732=0.267\]
Therefore, the exact value of \[\tan {{15}^{\circ }}\] using half angle formula is 0.267.
Note: Students make mistakes while applying the sine and cosine 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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

