1. Could you guys help me to manipulate this equation (half angle formulae) so that the subject would be tan B/2?
Equation: tan B = (2 tan B/2) / (1 - tan^2 A/2)
2. Express cos 6Q + cos 2Q as a product of trigonometric functions. Hence solve the equation cos2Q + cos4Q + cos 6Q = 0 for range of Q 0deg - 180 deg inclusive of 0 and 180.
(Ans: 2cos4Qcos2Q; 22.5deg, 60deg, 67.5deg, 112.5deg, 120deg, 157.5deg)
3. Prove (sin3A - sinA / cosA + cos3A) = tanA
THANKS! :D
Originally posted by Nathpoop:1. Could you guys help me to manipulate this equation (half angle formulae) so that the subject would be tan B/2?
Equation: tan B = (2 tan B/2) / (1 - tan^2 A/2)
2. Express cos 6Q + cos 2Q as a product of trigonometric functions. Hence solve the equation cos2Q + cos4Q + cos 6Q = 0 for range of Q 0deg - 180 deg inclusive of 0 and 180.
(Ans: 2cos4Qcos2Q; 22.5deg, 60deg, 67.5deg, 112.5deg, 120deg, 157.5deg)
3. Prove (sin3A - sinA / cosA + cos3A) = tanA
THANKS! :D
Hi Nathpoop,
For Question 1, tan B = tan (B/2 + B/2)
= [tan (B/2) + tan (B/2)] / [1 - tan(B/2)tan(B/2)]
= [2 tan (B/2)] / [1 - (tan (B/2))^2]
For Question 2, cos6Q +cos2Q = 2 X cos [(6Q + 2Q)/2] X cos [(6Q - 2Q)/2]
= 2(cos4Q)(cos2Q)
cos2Q + cos4Q + cos6Q = cos4Q + cos6Q + cos2Q
0 = cos4Q + 2(cos4Q)(cos2Q)
0 = (cos4Q)(1 + 2cos2Q)
cos4Q = 0 or cos2Q = -1/2
I believe that you will be able to solve from here.
For Question 3,
(sin3A - sinA) / (cosA + cos3A)
= 2 [(cos[(3A + A)/2])(sin[(3A - A)/2]] / 2 [(cos[(3A + A)/2])(cos[(3A - A)/2]]
= (sin[(3A - A)/2]] / (cos[(3A - A)/2
= sinA / cosA
= tanA
Cheers.