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Trigonometric Functions

Multiple Choice Questions

81.

If cot(α) and cot(β) are the roots of the equation x² – 3x + 2 = 0, then what is cot (α + β) equal to ?

$\frac{1}{2}$
$\frac{1}{3}$
2
3

82.

If sin(x) + sin(y) = cos(y) – cos(x), where $0 < y < x < \frac{π}{2}$

then what is $\tan (\frac{x - y}{2})$ equal to ?

0
$\frac{1}{2}$
1
2

83.

$Let α = β = 15 °$

What is sin(α + 1^o) + cos(β + 1^o) equal to ?

$\sqrt{3} \cos (1 °) + \sin (1 °)$
$\sqrt{3} \cos (1 °) - \frac{1}{2} \sin (1 °)$
$\frac{1}{\sqrt{2}} (\sqrt{3} \cos (1 °) - \sin (1 °))$
$\frac{1}{\sqrt{2}} (\cos (1 °) - \sin (1 °))$

84.

$Let α = β = 15 °$

What is the value of sin(7α) -cos(7β) ?

$\frac{1}{\sqrt{2}}$
$\frac{1}{2 \sqrt{2}}$
$\frac{\sqrt{3}}{2 \sqrt{2}}$
$\frac{\sqrt{3}}{\sqrt{2}}$

85.

$Let α = β = 15 °$

What is the value of sin(α) + cos(β) ?

$\frac{1}{\sqrt{2}}$
$\frac{1}{2 \sqrt{2}}$
$\frac{\sqrt{3}}{2 \sqrt{2}}$
$\frac{\sqrt{3}}{\sqrt{2}}$

86.

Let t_n = sinⁿ(θ) + cosⁿ(θ)

What is the value of t₁₀ where θ = 45^o ?

1
$\frac{1}{4}$
$\frac{1}{16}$
$\frac{1}{32}$

87.

Let t_n = sinⁿ(θ) + cosⁿ(θ)

What is the value of t₁² - t₂ = ?

$\cos (2 θ)$
$\sin (2 θ)$
$2 \cos (θ)$
$2 \sin (θ)$

88.

Let t_n = sinⁿ(θ) + cosⁿ(θ)

What is $\frac{t_{3} - t_{5}}{t_{5} - t_{7}}$ = ?

$\frac{t_{1}}{t_{3}}$
$\frac{t_{3}}{t_{5}}$
$\frac{t_{5}}{t_{7}}$
$\frac{t_{1}}{t_{7}}$

89.
Let asin²(x) + bcos²(x) = c; bsin²(y) + acos²(y) = d and p tan(x) = qtan(y).
What is $\frac{p^{2}}{q^{2}}$ equal to?

$\frac{(b - c) (b - d)}{(a - d) (a - c)}$

$\frac{(a - d) (c - a)}{(b - c) (d - b)}$

$\frac{(d - a) (c - a)}{(b - c) (d - b)}$

$\frac{(b - c) (b - d)}{(c - a) (a - d)}$

$\frac{(a - d) (c - a)}{(b - c) (d - b)}$

$ptan (x) = qtan (y) \Rightarrow \frac{p}{q} = \frac{\tan (y)}{\tan (x)} \Rightarrow \frac{p^{2}}{q^{2}} = \frac{\tan^{2} (y)}{\tan^{2} (x)} . . . (i) Now, {asin}^{2} (x) + {bcos}^{2} (x) = c . . . (ii) From (ii) \Rightarrow {asin}^{2} (x) + b (1 - \sin^{2} (x)) = c \Rightarrow {asin}^{2} (x) + b - {bsin}^{2} (x) = c \Rightarrow \sin^{2} (x) (a - b) = c - b \Rightarrow \sin^{2} (x) = \frac{c - b}{a - b} Also \Rightarrow a (1 - \cos^{2} (x)) + b (\cos^{2} (x)) = c \Rightarrow a - {acos}^{2} (x) + {bcos}^{2} (x) = c \Rightarrow \cos^{2} (x) = \frac{c - a}{b - a} \tan^{2} (x) = \frac{\sin^{2} (x)}{\cos^{2} (x)} = \frac{\frac{c - b}{a - b}}{\frac{c - a}{b - a}} = \frac{c - b}{a - c} . . . (iii) Consider {bsin}^{2} (y) + {acos}^{2} (y) = d . . . (iv) From (iv) \Rightarrow {bsin}^{2} (y) + a (1 - \sin^{2} (y)) = d \Rightarrow {bsin}^{2} (y) + a - {asin}^{2} (y) = d \Rightarrow \sin^{2} (y) (b - a) = d - a \Rightarrow \sin^{2} (y) = \frac{d - a}{b - a} Also \Rightarrow b (1 - \cos^{2} (y)) + {acos}^{2} (y) = d \Rightarrow b - {bcos}^{2} (y) + {asin}^{2} (y) = d \Rightarrow \cos^{2} (y) = \frac{d - b}{a - b} Divide by (ii) by (iii) \frac{d - a}{b - d} = \tan^{2} (y) . . . (v)$