How do you use the half angle formula to determine the exact values of the sines, cosine, and tangent $\dfrac{19\pi }{12}$ ?
Answer
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Hint: In this question, we have to find the exact value of trigonometric functions. Thus, we will use the half angle formula to get the solution. As we know, the half angle formula allows the angles of trigonometric functions equal to $\dfrac{x}{2}$ in terms of x. Thus, in this problem, we will first find the value of x, by equating $\dfrac{x}{2}=\dfrac{19\pi }{12}$ . After that, we will apply the half angle formula of sin function, that is $\sin \left( \dfrac{x}{2} \right)=\pm \sqrt{\dfrac{1-\cos x}{2}}$ . The, we will find the value cos function, where $\cos \left( \dfrac{x}{2} \right)=\pm \sqrt{\dfrac{1+\cos x}{2}}$ . As we know, tan function is the division of sine and cosine function, thus we will apply the formula $\tan \left( \dfrac{x}{2} \right)=\dfrac{\sin \left( \dfrac{x}{2} \right)}{\cos \left( \dfrac{x}{2} \right)}$ , to get all the required solution for the problem.
Complete step by step answer:
According to the problem, we have to find the exact value of trigonometric functions.
Thus, we will apply the half angle formula to get the solution.
The angle given to us is $\dfrac{19\pi }{12}$ ---------- (1)
Let us now, first find the value of x, by equating $\dfrac{x}{2}$ and value of equation (1), we get
$\Rightarrow \dfrac{x}{2}=\dfrac{19\pi }{12}$
Now, we will multiply 2 on both sides in the above equation, we get
$\Rightarrow \dfrac{x}{2}\times 2=\dfrac{19\pi }{12}\times 2$
On further solving the above equation, we get
$\Rightarrow x=\dfrac{19\pi }{6}$ --------- (2)
Now, we know that in the half angle formula, we have one trigonometric function which is equal to $\cos x$ , thus we will now find the value of the same function, that is
$\cos x=\cos \dfrac{19\pi }{6}$
$\Rightarrow \cos x=\cos \left( 2\pi +\dfrac{7\pi }{6} \right)$
Now, we know that any angle which lies after $2\pi $ , lies in the first quadrant, hence all the trigonometric functions are positive, therefore $\cos \left( 2\pi +x \right)=\cos x$ , we get
$\Rightarrow \cos \left( \dfrac{7\pi }{6} \right)$
$\Rightarrow \cos \left( \pi +\dfrac{\pi }{6} \right)$
Again, we know that the angle after $\pi $ lies after third quadrant, hence the cos function is negative, which implies $\cos \left( \pi +x \right)=-\cos x$
$\Rightarrow -\cos \left( \dfrac{\pi }{6} \right)$
$\Rightarrow \cos x=\cos \dfrac{19\pi }{6}=-\dfrac{\sqrt{3}}{2}$ ---------- (3)
So, now we will find the value of sine function using half-angle formula, such that $\sin \left( \dfrac{x}{2} \right)=\pm \sqrt{\dfrac{1-\cos x}{2}}$ , thus we will substitute the value of equation (1) in the given formula, we get
$\Rightarrow \sin \left( \dfrac{19\pi }{12} \right)=\pm \sqrt{\dfrac{1-\cos \left( \dfrac{19\pi }{6} \right)}{2}}$
Thus, now we will substitute the value of equation (3) in the above equation, we get
$\Rightarrow \pm \sqrt{\dfrac{1-\left( \dfrac{-\sqrt{3}}{2} \right)}{2}}$
Now, taking the LCM in the above value, we get
$\Rightarrow \pm \sqrt{\dfrac{\dfrac{2-\left( -\sqrt{3} \right)}{2}}{2}}$
On further solving the above equation, we get
$\Rightarrow \pm \sqrt{\dfrac{2+\sqrt{3}}{4}}$
Also, we know that $\sin \left( \dfrac{19\pi }{12} \right)$ lies in fourth quadrant, thus we know that sin function is negative in the same quadrant, therefore
$\Rightarrow \sin \left( \dfrac{19\pi }{12} \right)=-\sqrt{\dfrac{2+\sqrt{3}}{4}}$ ----------- (4)
Similarly, we will find the value of cosine function, that is the two angle formula of sine function is equal to $\cos \left( \dfrac{x}{2} \right)=\pm \sqrt{\dfrac{1+\cos x}{2}}$ , thus we will substitute the value of equation (1) in the given formula, we get
$\Rightarrow \cos \left( \dfrac{19\pi }{12} \right)=\pm \sqrt{\dfrac{1+\cos \left( \dfrac{19\pi }{6} \right)}{2}}$
Thus, now we will substitute the value of equation (3) in the above equation, we get
$\Rightarrow \pm \sqrt{\dfrac{1+\left( \dfrac{-\sqrt{3}}{2} \right)}{2}}$
Now, taking the LCM in the above value, we get
$\Rightarrow \pm \sqrt{\dfrac{\dfrac{2+\left( -\sqrt{3} \right)}{2}}{2}}$
On further solving the above equation, we get
$\Rightarrow \pm \sqrt{\dfrac{2-\sqrt{3}}{4}}$
Also, we know that $\cos \left( \dfrac{19\pi }{12} \right)$ lies in fourth quadrant, thus we know that cos function is positive in the same quadrant, therefore
$\Rightarrow \cos \left( \dfrac{19\pi }{12} \right)=\sqrt{\dfrac{2-\sqrt{3}}{4}}$ ------------- (5)
Also, we know that the tan function is equal to the division of the sin function and the cos function, that is $\tan \left( \dfrac{x}{2} \right)=\dfrac{\sin \left( \dfrac{x}{2} \right)}{\cos \left( \dfrac{x}{2} \right)}$ . Thus, we will substitute the value (4) and (5) in the given formula, we get
$\Rightarrow \tan \left( \dfrac{x}{2} \right)=\dfrac{-\sqrt{\dfrac{2+\sqrt{3}}{4}}}{\sqrt{\dfrac{2-\sqrt{3}}{4}}}$
On further simplification, we get
$\Rightarrow \tan \left( \dfrac{x}{2} \right)=\dfrac{-\sqrt{2+\sqrt{3}}}{\sqrt{2-\sqrt{3}}}$
Now, we will rationalize the denominator by multiplying and divide $\sqrt{2-\sqrt{3}}$ on the right hand side of the equation, we get
$\Rightarrow \tan \left( \dfrac{19\pi }{12} \right)=\dfrac{-\sqrt{2+\sqrt{3}}}{\sqrt{2-\sqrt{3}}}\times \dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{2-\sqrt{3}}}$
Now, we will apply the formula $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$ and $\sqrt{a}\times \sqrt{a}=a$ in the above equation, we get
$\Rightarrow \tan \left( \dfrac{19\pi }{12} \right)=\dfrac{-\sqrt{\left( 2+\sqrt{3} \right)\left( 2-\sqrt{3} \right)}}{2-\sqrt{3}}$
Now, we will apply the algebraic identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ in the above equation, we get
$\Rightarrow \tan \left( \dfrac{19\pi }{12} \right)=\dfrac{-\sqrt{{{\left( 2 \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}}{2-\sqrt{3}}$
Therefore, we get
$\Rightarrow \tan \left( \dfrac{19\pi }{12} \right)=\dfrac{-\sqrt{4-3}}{2-\sqrt{3}}$
$\Rightarrow \tan \left( \dfrac{19\pi }{12} \right)=\dfrac{-1}{2-\sqrt{3}}$
Therefore, for the angle $\dfrac{19\pi }{12}$ , sine function is equal to $-\sqrt{\dfrac{2+\sqrt{3}}{4}}$ , cosine function is equal to $\sqrt{\dfrac{2-\sqrt{3}}{4}}$ , and the tan function is equal to $\dfrac{-1}{2-\sqrt{3}}$ .
Note:
While solving this problem, do mention all the steps properly to avoid error and confusion. Using a two-angle formula, you have to find the value of x and then solve further to get the required solution.
Complete step by step answer:
According to the problem, we have to find the exact value of trigonometric functions.
Thus, we will apply the half angle formula to get the solution.
The angle given to us is $\dfrac{19\pi }{12}$ ---------- (1)
Let us now, first find the value of x, by equating $\dfrac{x}{2}$ and value of equation (1), we get
$\Rightarrow \dfrac{x}{2}=\dfrac{19\pi }{12}$
Now, we will multiply 2 on both sides in the above equation, we get
$\Rightarrow \dfrac{x}{2}\times 2=\dfrac{19\pi }{12}\times 2$
On further solving the above equation, we get
$\Rightarrow x=\dfrac{19\pi }{6}$ --------- (2)
Now, we know that in the half angle formula, we have one trigonometric function which is equal to $\cos x$ , thus we will now find the value of the same function, that is
$\cos x=\cos \dfrac{19\pi }{6}$
$\Rightarrow \cos x=\cos \left( 2\pi +\dfrac{7\pi }{6} \right)$
Now, we know that any angle which lies after $2\pi $ , lies in the first quadrant, hence all the trigonometric functions are positive, therefore $\cos \left( 2\pi +x \right)=\cos x$ , we get
$\Rightarrow \cos \left( \dfrac{7\pi }{6} \right)$
$\Rightarrow \cos \left( \pi +\dfrac{\pi }{6} \right)$
Again, we know that the angle after $\pi $ lies after third quadrant, hence the cos function is negative, which implies $\cos \left( \pi +x \right)=-\cos x$
$\Rightarrow -\cos \left( \dfrac{\pi }{6} \right)$
$\Rightarrow \cos x=\cos \dfrac{19\pi }{6}=-\dfrac{\sqrt{3}}{2}$ ---------- (3)
So, now we will find the value of sine function using half-angle formula, such that $\sin \left( \dfrac{x}{2} \right)=\pm \sqrt{\dfrac{1-\cos x}{2}}$ , thus we will substitute the value of equation (1) in the given formula, we get
$\Rightarrow \sin \left( \dfrac{19\pi }{12} \right)=\pm \sqrt{\dfrac{1-\cos \left( \dfrac{19\pi }{6} \right)}{2}}$
Thus, now we will substitute the value of equation (3) in the above equation, we get
$\Rightarrow \pm \sqrt{\dfrac{1-\left( \dfrac{-\sqrt{3}}{2} \right)}{2}}$
Now, taking the LCM in the above value, we get
$\Rightarrow \pm \sqrt{\dfrac{\dfrac{2-\left( -\sqrt{3} \right)}{2}}{2}}$
On further solving the above equation, we get
$\Rightarrow \pm \sqrt{\dfrac{2+\sqrt{3}}{4}}$
Also, we know that $\sin \left( \dfrac{19\pi }{12} \right)$ lies in fourth quadrant, thus we know that sin function is negative in the same quadrant, therefore
$\Rightarrow \sin \left( \dfrac{19\pi }{12} \right)=-\sqrt{\dfrac{2+\sqrt{3}}{4}}$ ----------- (4)
Similarly, we will find the value of cosine function, that is the two angle formula of sine function is equal to $\cos \left( \dfrac{x}{2} \right)=\pm \sqrt{\dfrac{1+\cos x}{2}}$ , thus we will substitute the value of equation (1) in the given formula, we get
$\Rightarrow \cos \left( \dfrac{19\pi }{12} \right)=\pm \sqrt{\dfrac{1+\cos \left( \dfrac{19\pi }{6} \right)}{2}}$
Thus, now we will substitute the value of equation (3) in the above equation, we get
$\Rightarrow \pm \sqrt{\dfrac{1+\left( \dfrac{-\sqrt{3}}{2} \right)}{2}}$
Now, taking the LCM in the above value, we get
$\Rightarrow \pm \sqrt{\dfrac{\dfrac{2+\left( -\sqrt{3} \right)}{2}}{2}}$
On further solving the above equation, we get
$\Rightarrow \pm \sqrt{\dfrac{2-\sqrt{3}}{4}}$
Also, we know that $\cos \left( \dfrac{19\pi }{12} \right)$ lies in fourth quadrant, thus we know that cos function is positive in the same quadrant, therefore
$\Rightarrow \cos \left( \dfrac{19\pi }{12} \right)=\sqrt{\dfrac{2-\sqrt{3}}{4}}$ ------------- (5)
Also, we know that the tan function is equal to the division of the sin function and the cos function, that is $\tan \left( \dfrac{x}{2} \right)=\dfrac{\sin \left( \dfrac{x}{2} \right)}{\cos \left( \dfrac{x}{2} \right)}$ . Thus, we will substitute the value (4) and (5) in the given formula, we get
$\Rightarrow \tan \left( \dfrac{x}{2} \right)=\dfrac{-\sqrt{\dfrac{2+\sqrt{3}}{4}}}{\sqrt{\dfrac{2-\sqrt{3}}{4}}}$
On further simplification, we get
$\Rightarrow \tan \left( \dfrac{x}{2} \right)=\dfrac{-\sqrt{2+\sqrt{3}}}{\sqrt{2-\sqrt{3}}}$
Now, we will rationalize the denominator by multiplying and divide $\sqrt{2-\sqrt{3}}$ on the right hand side of the equation, we get
$\Rightarrow \tan \left( \dfrac{19\pi }{12} \right)=\dfrac{-\sqrt{2+\sqrt{3}}}{\sqrt{2-\sqrt{3}}}\times \dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{2-\sqrt{3}}}$
Now, we will apply the formula $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$ and $\sqrt{a}\times \sqrt{a}=a$ in the above equation, we get
$\Rightarrow \tan \left( \dfrac{19\pi }{12} \right)=\dfrac{-\sqrt{\left( 2+\sqrt{3} \right)\left( 2-\sqrt{3} \right)}}{2-\sqrt{3}}$
Now, we will apply the algebraic identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ in the above equation, we get
$\Rightarrow \tan \left( \dfrac{19\pi }{12} \right)=\dfrac{-\sqrt{{{\left( 2 \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}}{2-\sqrt{3}}$
Therefore, we get
$\Rightarrow \tan \left( \dfrac{19\pi }{12} \right)=\dfrac{-\sqrt{4-3}}{2-\sqrt{3}}$
$\Rightarrow \tan \left( \dfrac{19\pi }{12} \right)=\dfrac{-1}{2-\sqrt{3}}$
Therefore, for the angle $\dfrac{19\pi }{12}$ , sine function is equal to $-\sqrt{\dfrac{2+\sqrt{3}}{4}}$ , cosine function is equal to $\sqrt{\dfrac{2-\sqrt{3}}{4}}$ , and the tan function is equal to $\dfrac{-1}{2-\sqrt{3}}$ .
Note:
While solving this problem, do mention all the steps properly to avoid error and confusion. Using a two-angle formula, you have to find the value of x and then solve further to get the required solution.
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