Ravi Kant Kumar

If cos x = -3/5 and x lies in the IInd quadrant, find sin2x and sin(x/2) If \[ \cos x=-\frac35 \] and \(x\) lies in the IInd quadrant, find the values of \[ \sin2x \quad \text{and} \quad \sin\frac{x}{2} \] Solution: Given, \[ \cos x=-\frac35 \] Using \[ \sin^2x+\cos^2x=1 \] we get \[ \sin^2x = 1-\left(-\frac35\right)^2

If cos x = -3/5 and x lies in the IIIrd quadrant, find cos(x/2), sin(x/2) and sin2x If \[ \cos x=-\frac35 \] and \(x\) lies in the IIIrd quadrant, find the values of \[ \cos\frac{x}{2},\quad \sin\frac{x}{2},\quad \text{and} \quad \sin2x \] Solution: Given, \[ \cos x=-\frac35 \] Using \[ \sin^2x+\cos^2x=1 \] we get \[ \sin^2x =

Prove that cot(π/8) = √2 + 1 Prove that \[ \cot\frac{\pi}{8}=\sqrt2+1 \] Proof: Using the half-angle identity: \[ \tan\frac{\theta}{2} = \frac{1-\cos\theta}{\sin\theta} \] Let \[ \theta=\frac{\pi}{4} \] Then, \[ \tan\frac{\pi}{8} = \frac{1-\cos\frac{\pi}{4}}{\sin\frac{\pi}{4}} \] Using standard values: \[ \sin\frac{\pi}{4}=\frac{\sqrt2}{2} \] and \[ \cos\frac{\pi}{4}=\frac{\sqrt2}{2} \] Substituting: \[ \tan\frac{\pi}{8} = \frac{1-\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}} \] \[ = \frac{2-\sqrt2}{\sqrt2} \] \[ = \sqrt2-1

Prove that tan 82½° = (√3 + √2)(√2 + 1) Prove that \[ \tan82\frac12^\circ = (\sqrt3+\sqrt2)(\sqrt2+1) = \sqrt2+\sqrt3+\sqrt4+\sqrt6 \] Proof: \[ 82\frac12^\circ = 45^\circ+37\frac12^\circ \] Also, \[ 37\frac12^\circ=\frac{75^\circ}{2} \] Using the identity \[ \tan\left(45^\circ+\theta\right) = \frac{1+\tan\theta}{1-\tan\theta} \] let \[ \theta=37\frac12^\circ \] Then, \[ \tan82\frac12^\circ = \frac{1+\tan37\frac12^\circ}{1-\tan37\frac12^\circ} \] Using the half-angle identity \[ \tan\frac{\theta}{2} = \frac{\sin\theta}{1+\cos\theta}

Prove that sin3x + sin2x − sinx = 4sinx cos(x/2) cos(3x/2) Prove that \[ \sin3x+\sin2x-\sin x = 4\sin x\cos\frac{x}{2}\cos\frac{3x}{2} \] Proof: Start with the left-hand side: \[ LHS=\sin3x+\sin2x-\sin x \] Group the last two terms: \[ =\sin3x+(\sin2x-\sin x) \] Using the identity \[ \sin A-\sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \] with \[ A=2x,\quad B=x \] we

Prove that cos4x − cos4α = 8(cosx − cosα)(cosx + cosα)(cosx − sinα)(cosx + sinα) Prove that \[ \cos4x-\cos4\alpha = 8(\cos x-\cos\alpha) (\cos x+\cos\alpha) (\cos x-\sin\alpha) (\cos x+\sin\alpha) \] Proof: Start with the left-hand side: \[ LHS=\cos4x-\cos4\alpha \] Using \[ \cos4\theta=8\cos^4\theta-8\cos^2\theta+1 \] we get \[ \cos4x = 8\cos^4x-8\cos^2x+1 \] and \[ \cos4\alpha = 8\cos^4\alpha-8\cos^2\alpha+1 \]

Prove that cot²x − tan²x = 4cot2x cosec2x Prove that \[ \cot^2x-\tan^2x=4\cot2x\cosec2x \] Proof: Start with the left-hand side: \[ LHS=\cot^2x-\tan^2x \] Convert into sine and cosine form: \[ =\frac{\cos^2x}{\sin^2x}-\frac{\sin^2x}{\cos^2x} \] Taking LCM: \[ =\frac{\cos^4x-\sin^4x}{\sin^2x\cos^2x} \] Using \[ a^4-b^4=(a^2-b^2)(a^2+b^2) \] we get \[ =\frac{(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)}{\sin^2x\cos^2x} \] Using \[ \cos^2x+\sin^2x=1 \] and \[ \cos2x=\cos^2x-\sin^2x \] therefore, \[

Prove that tan(π/4 + x) + tan(π/4 − x) = 2sec2x Prove that \[ \tan\left(\frac{\pi}{4}+x\right) + \tan\left(\frac{\pi}{4}-x\right) = 2\sec2x \] Proof: Using the identity \[ \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \] we get \[ \tan\left(\frac{\pi}{4}+x\right) = \frac{1+\tan x}{1-\tan x} \] Similarly, \[ \tan\left(\frac{\pi}{4}-x\right) = \frac{1-\tan x}{1+\tan x} \] Therefore, \[ LHS = \frac{1+\tan x}{1-\tan x}

Prove that cos⁶x − sin⁶x = cos2x(1 − ¼sin²2x) Prove that \[ \cos^6x-\sin^6x = \cos2x\left(1-\frac14\sin^22x\right) \] Proof: Start with the left-hand side: \[ LHS=\cos^6x-\sin^6x \] Using the identity \[ a^3-b^3=(a-b)(a^2+ab+b^2) \] with \[ a=\cos^2x,\quad b=\sin^2x \] we get \[ LHS = (\cos^2x-\sin^2x) (\cos^4x+\cos^2x\sin^2x+\sin^4x) \] Using \[ \cos2x=\cos^2x-\sin^2x \] therefore, \[ LHS = \cos2x (\cos^4x+\cos^2x\sin^2x+\sin^4x) \]

Prove that 2(sin⁶x + cos⁶x) − 3(sin⁴x + cos⁴x) + 1 = 0 Prove that \[ 2(\sin^6x+\cos^6x) – 3(\sin^4x+\cos^4x) +1 =0 \] Proof: Using \[ a^3+b^3=(a+b)^3-3ab(a+b) \] with \[ a=\sin^2x,\quad b=\cos^2x \] we get \[ \sin^6x+\cos^6x = (\sin^2x+\cos^2x)^3 – 3\sin^2x\cos^2x(\sin^2x+\cos^2x) \] Using \[ \sin^2x+\cos^2x=1 \] therefore, \[ \sin^6x+\cos^6x = 1-3\sin^2x\cos^2x \] Also, \[ \sin^4x+\cos^4x =