Trigonometric Identities

Question

If sin(theta) = 0.3 and 0 < theta < pi / 2 ( first quadrant ). Find each of the following :

A) sin (2theta)

B) sin (4theta)

C) cos (4theta)

D) sin (theta/2)

E) cos (theta/2)

———————————————————————————

Hello! I am having difficulty understanding where to go beyond where I left off. Starting with the first letter A:

I obtained each sides from what was given (.3 being our opposite and 1 as our hypotenuse), then did the Pythagorean theorem and got √.91 as the adjacent side.

Following shows my work where I don’t know how to locate these points on the unit circle or learned an alternative.

Sin 2theta —> 2sinucosu

2sin(.3)cos√.91

If there is anyway to understand better please let me know or if I am on the right track. Thank you!

Answer 1

sinT=.3=3/10

cosT =sqr(10^2-3^2)/10 =sqr91/10=0.954

sin2T = 2sinTcosT = 2(.3)sqr91/10 = .06sqr91 = about 0.57

T = sin^-1(.3) = about 17.46 degrees

2T = 34.92

sin2T = sin34.92 = about 0.57

sin4T = 2sin2Tcos2T = 2(0.57)(.82)= about 0.93

sin69.84 = about 0.94

cos4T = cos69.84 = about 0.34

cos4T = cos^2(2T) - sin^2(2T) = (.82)^2 - (.57)^2 = about .67-.33 = 0.34

use half angle formulas for the last two

sinT/2 = sqr((1-cosT)/2)= sqr(1-.954)/2) = sqr(.046/2) = sqr(.023) = 0.152

cosT/2 = sqr((1+cosT)/2) = sqr(1.954/2) = sqr0.977 =0.988

and 17.4/2= 8.72 degrees

use a calculator with a trig function to check your answers

sin8.72 = .152

cos8.72= ,988