Value of the expression 2sinx – cos2x is

Q: Value of the expression 2sinx – cos2x is

(A) greater than -3/2

(B) equal to – 3/2

(C) less than -3/2

(D) none of these

Sol. 2 sinx – cos2x = 2 (sin2x + sinx) – 1

$\large = 2 [(sinx + \frac{1}{2})^2 – \frac{1}{4}] – 1 $

$\large = 2(sinx + \frac{1}{2})^2 – \frac{3}{2} \ge -\frac{3}{2}$

⇒ 2 sinx – cos2x ≥ – 3/2

Hence (A), (B) are the correct answers.

If Tn = sin^nθ + cos^nθ , then $\frac{T_6 – T_4}{T_6} = m $ holds for values of m satisfying

Q: If Tn = sinnθ + cosnθ , then $\frac{T_6 – T_4}{T_6} = m $ holds for values of m satisfying

(A) $\large m \in [-1 , \frac{1}{3}]$

(B) $\large m \in [0 , \frac{1}{3}]$

(C) $\large m \in [-1 , 0]$

(D) $\large m \in [-1 , -\frac{1}{2}]$

Sol: $\large \frac{T_6 – T_4}{T_6} = \frac{-sin^2 2\theta}{4-3sin^2 2\theta} = m $

$\large sin^2 2\theta = \frac{4 m}{-1 + 3 m}$

$\large 0 \le \frac{4 m}{-1 + 3 m} \le 1 $

$\large m \in [-1 , 0]$

Hence (C), (D) are the correct answers

If 3 sinβ = sin(2α+ β), then tan(α+ β) is equal to

Q: If 3 sinβ = sin(2α+ β), then tan(α+ β) is equal to

(A) 2 tanβ

(B) 2 tanα

(C) tanα + tanβ

(D) none of these

Sol. $\large \frac{sin(2\alpha + \beta)}{sin\beta} = \frac{3}{1}$

$\large \frac{sin(2\alpha + \beta) + sin\beta}{sin(2\alpha + \beta) – sin\beta} = \frac{3+1}{3-1} = 2 $

$\large \frac{2 sin(\alpha + \beta) cos\alpha}{2sin\alpha cos(\alpha + \beta)} = 2$

⇒ tan(α + β) = 2 tanα

Hence (B) is the correct answer.