It is known that
\(2\,cosA+cosB=2cos(\frac{A+B}{2})cos(\frac{A-B}{2}),\,cosA-cosB=2sin(\frac{A+B}{2})sin(\frac{A-B}{2})\)
∴L.H.S. = cos2 2x-cos2 6x
= (cos 2x +cos 6x) (cos 2x-6x)
\(=[2\,cos(\frac{2x+6x}{2})cos(\frac{2x-6x}{2})][2\,cos(\frac{2x+6x}{2})sin(\frac{2x-6x}{2})]\)
= [2 cos 4x cos (-2x)] [-2 sin 4x (-sin 2x)]
= (2 sin 4x cos 4x) (2 sin 2x cos 2x)
= sin 8x sin 4x
= R.H.S
Prove that. \(sin^2 \frac{π}{6}+cos^2 \frac{π}{3}–tan^2 \frac{π}{4}=-\frac{1}{2}\)
Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?
The relationship between the sides and angles of a right-angle triangle is described by trigonometry functions, sometimes known as circular functions. These trigonometric functions derive the relationship between the angles and sides of a triangle. In trigonometry, there are three primary functions of sine (sin), cosine (cos), tangent (tan). The other three main functions can be derived from the primary functions as cotangent (cot), secant (sec), and cosecant (cosec).
sin x = a/h
cos x = b/h
tan x = a/b
Tan x can also be represented as sin x/cos x
sec x = 1/cosx = h/b
cosec x = 1/sinx = h/a
cot x = 1/tan x = b/a