If c o s ( θ − α ) = a, s i n ( θ − β ) = b, then c o s 2 ( α − β ) + 2ab s i n ( α − β ) is equal to Options: A . &nbsp 4 a 2 b 2 B . &nbsp a 2 − b 2 C . &nbsp a 2 + b 2 D . &nbsp - a 2 b 2

Answer: Option C : C We have s i n ( α − β ) = s i n ( θ − β − ¯ ¯¯¯¯¯¯¯¯¯¯¯ ¯ θ − α ) = s i n ( θ − β ) c o s ( θ − α ) − c o s ( θ − β ) s i n ( θ − α ) = ba - √ 1 − b 2 √ 1 − a 2 And c o s ( α − β ) = c o s ( θ − β − ¯ ¯¯¯¯¯¯¯¯¯¯¯ ¯ θ − α ) = c o s ( θ − β ) c o s ( θ − α ) + s i n ( θ − β ) s i n ( θ − α ) = a √ 1 − b 2 + b √ 1 − a 2 ∴ Given expression is c o s 2 ( α − β ) + 2 a b s i n ( α − β ) = ( a √ 1 − b 2 + b √ 1 − a 2 ) 2 + 2ab{ a b − √ 1 − a 2 √ 1 − b 2 } = a 2 + b 2 . Trick: Put α = 30 ∘ , β = 60 ∘ and θ = 90 ∘ , then a = 1 2 , b = 1 2 ∴ c o s 2 ( α − β ) + 2 a b s i n ( α − β ) = 3 4 + 1 2 × (- 1 2 ) = 1 2 which is given by option (c).

If cos(θ−α) = a, sin(θ−β) = b, then cos

Discussion Forum : Trigonometric Functions

Question -

If $c o s (θ - α)$ = a, $s i n (θ - β)$ = b,

then $c o s^{2} (α - β)$ + 2ab $s i n (α - β)$ is equal to

Options:

A .

4 a^{2} b^{2}

B .

a^{2} - b^{2}

C .

a^{2} + b^{2}

D . -

a^{2} b^{2}

Answer: Option C
:
C

We have $s i n (α - β)$ = $s i n (θ - β - ¯ ¯¯¯¯¯¯¯¯¯¯¯ ¯ θ - α)$

= $s i n (θ - β) c o s (θ - α) - c o s (θ - β) s i n (θ - α)$

= ba - $\sqrt{1 - b^{2}} \sqrt{1 - a^{2}}$

And $c o s (α - β) = c o s (θ - β - ¯ ¯¯¯¯¯¯¯¯¯¯¯ ¯ θ - α)$

= $c o s (θ - β) c o s (θ - α) + s i n (θ - β) s i n (θ - α)$

= $a \sqrt{1 - b^{2}} + b \sqrt{1 - a^{2}}$

∴ Given expression is $c o s^{2} (α - β) + 2 a b s i n (α - β)$

= $(a \sqrt{1 - b^{2}} + b \sqrt{1 - a^{2}})^{2}$ + 2ab{ $a b - \sqrt{1 - a^{2}} \sqrt{1 - b^{2}}$ }

= $a^{2} + b^{2}$ .

Trick: Put $α = 30^{\circ}$ , $β = 60^{\circ}$ and $θ = 90^{\circ}$ ,

then a = $\frac{1}{2}$ , b = $\frac{1}{2}$

∴ $c o s^{2} (α - β) + 2 a b s i n (α - β)$ = $\frac{3}{4}$ + $\frac{1}{2}$ × (- $\frac{1}{2}$ ) = $\frac{1}{2}$

which is given by option (c).

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