Question

How do you find the exact values of \[\sin 2u,\ \cos 2u,\ \tan 2u\] using the double angle values given \[\cos u=\dfrac{-2}{3}\] ,\[\dfrac{\pi }{2} \] < u < \[\pi \] ?

Answer 1

Hint: In order to find the exact values, first by using the unit circle right triangle and Pythagoras theorem we will find the other third side of the side as the two sides of the triangle are given in the form of \[\cos u=\dfrac{base}{hypotenuse}=-\dfrac{2}{3}\] . Later using the double angles formula of sine, cosine and the tangent function and solving the expression by substituting the values, we will get the exact values of \[\sin 2u,\ \cos 2u,\ \tan 2u\] .

Complete step-by-step answer:
We have given that,
 \[\cos u=-\dfrac{2}{3}\]
And the value of ‘u’ lies between \[\dfrac{\pi }{2}The given implies that ‘u’ lies on the second quadrant.
In the second quadrant the value of sin is positive only and the value of cosine function is negative.
Now,
Using the definition of cos function to find the known sides of the unit circle right triangle. The quadrant determines the sign on each of the values.
 \[\cos u=\dfrac{base}{hypotenuse}=-\dfrac{2}{3}\]
Finding the perpendicular side of the unit circle triangle. Since we know the values of base and the hypotenuse.
Using the Pythagoras theorem, i.e.
 \[{{\left( hypotenuse \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}\]
 \[perpendicular=\sqrt{{{\left( hypotenuse \right)}^{2}}-{{\left( base \right)}^{2}}}\]
Putting the values, we get
 \[perpendicular=\sqrt{{{\left( 3 \right)}^{2}}-{{\left( 2 \right)}^{2}}}=\sqrt{9-4}=\sqrt{5}\]
Using the definition of sine function,
 \[\sin u=\dfrac{perpendicular}{hypotenuse}=\dfrac{\sqrt{5}}{3}\]
Therefore,
We have,
 \[\sin u=\dfrac{\sqrt{5}}{3}\] , \[\cos u=-\dfrac{2}{3}\]
Now,
Using the double angle formulas of trigonometric function;
 \[\sin 2u=2\sin u\cos u\]
 \[\Rightarrow \sin 2u=2\left( \dfrac{\sqrt{5}}{3} \right)\left( \dfrac{2}{3} \right)=\dfrac{4\sqrt{5}}{9}\]
 \[\Rightarrow \sin 2u=-\dfrac{4\sqrt{5}}{9}\]
As \[\dfrac{\pi }{2} \] < u < \[\pi \] then \[\pi \] < 2u <2 \[\pi \]
Thus, the value of 2u lies in the third quadrant. And in the third quadrant the value of sine function is negative.
 \[\cos 2u=2{{\cos }^{2}}u-1\]
 \[\Rightarrow \cos 2u=2{{\left( -\dfrac{2}{3} \right)}^{2}}-1=2\left( \dfrac{4}{9} \right)-1=\dfrac{8}{9}-1=-\dfrac{1}{9}\]
 \[\Rightarrow \cos 2u=-\dfrac{1}{9}\]
 \[\tan 2u=\dfrac{2\tan u}{1-{{\tan }^{2}}u}\]
Now,
 \[\tan u=\dfrac{\sin u}{\cos u}=\dfrac{\dfrac{\sqrt{5}}{3}}{-\dfrac{2}{3}}=\dfrac{\sqrt{5}}{3}\times -\dfrac{3}{2}=-\dfrac{\sqrt{5}}{2}\]
Therefore,
 \[\Rightarrow \tan 2u=\dfrac{2\tan u}{1-{{\tan }^{2}}u}=\dfrac{2\left( -\dfrac{\sqrt{5}}{2} \right)}{1-{{\left( -\dfrac{\sqrt{5}}{2} \right)}^{2}}}=\dfrac{-\sqrt{5}}{1-\dfrac{5}{4}}=\dfrac{-\sqrt{5}}{-\dfrac{1}{4}}=4\sqrt{5}\]
 \[\Rightarrow \tan 2u=4\sqrt{5}\]

Note: While solving these types of questions, students must need to remember the formulas of double angle of trigonometric functions. One must be careful while noted down the values to avoid any error in the answer. They also need to consider the quadrant which is given in the question and they must remember which trigonometric function is positive or negative in all the four quadrants. We must know the basic concept of trigonometry and also the definition of the trigonometric function. The sine, cosine and the tangent are the three basic functions in introduction to trigonometry which shows the relation between all the sides of the triangles.