# Prove 8. 2 / (√3 cos x + sin x) = sec((π / 6) - x)?

• Prove 8. 2 / (√3 cos x + sin x) = sec((π / 6) - x)?

Answer #1 | 29/04 2014 18:38
I'm not sure why you have 8.2 in numerator. It should only be 2. If 8 is the problem number, then it's really not necessary here and just confuses things. 2 / (√3 cos x + sin x) = 1 / (√3/2 cos x + 1/2 sin x) = 1 / (cos(π/6) cosx + sin(π/6) sinx) = 1 / cos(π/6 − x) = sec(π/6 − x)
Positive: 64 %
Answer #2 | 29/04 2014 19:06
cos(π/6 − x) = cos(π/6)cosx + sin(π/6)sinx) = [√3cos(x) + sin(x)]/2 sec(π/6 − x) = 1/cos(π/6 − x) = 2/[√3 cos(x) + sin(x)] It was probably question number 8.
Positive: 58 %

... pi }{8} \right )\left ( 1-cos^{2}\frac{3\pi }{8} \right ) = sin^{2 ... π/6) -1 = 1-4sin^2 (x+π/12)? If ... Prove that sec^2 x+cos^2 x>=2 for all x ...
Positive: 64 %
sin 2 (x) + cos 2 (x) = 1. tan 2 (x) + 1 = sec 2 (x) cot 2 (x) + 1 = csc 2 (x) sin(x y) = sin x cos y cos x sin y. cos(x y) = cos x cosy sin x sin y
Positive: 59 %
Basic Trig Identities – Sec. 7.1 . sin x ... 1 = – cos2 x cos 2 x – 1 = – sin 2 x . sec 2 x – tan 2 x = 1 ... Sec. 7.3 . cos (x ...
2 x" sin 2x # 1 8 sin 4x) # C! 1 4 y (3 2" 2 cos 2x # 1 ... 2 "1 " cos 2x# 2 TRIGONOMETRIC INTEGRALS ... ! y u6#1 "u2$du! y #u6 " u8$ du! y tan6x #1 ...