If cos θ = 3 5 and cos ϕ = 4 5 , where θ and ϕ are positive acute angles, then cos θ − ϕ 2 = Options: A . &nbsp 7 √ 2 B . &nbsp 7 5 √ 2 C . &nbsp 7 √ 5 D . &nbsp 7 2 √ 5

Answer: Option B : B We have cos θ = 3 5 and cos ϕ = 4 5 . Therefore c o s ( θ − ϕ ) = c o s θ c o s ϕ + s i n θ s i n ϕ = 3 5 . 4 5 + 4 5 . 3 5 = 24 25 But 2 c o s 2 ( θ − ϕ 2 ) = 1 + c o s ( θ − ϕ ) = 1 + 24 25 = 49 50 ∴ c o s 2 ( θ − ϕ 2 ) = 49 50 . Hence, c o s ( θ − ϕ 2 ) = 7 5 √ 2 .

If cos θ = 35 and cos ϕ = 45, where θ an

Discussion Forum : Trigonometric Functions

Question -

If cos $θ$ =
$\frac{3}{5}$ and cos $ϕ$ =
$\frac{4}{5}$ , where $θ$ and $ϕ$ are

positive acute angles, then cos
$\frac{θ - ϕ}{2}$ =

Options:

A .

\frac{7}{\sqrt{2}}

B .

\frac{7}{5 \sqrt{2}}

C .

\frac{7}{\sqrt{5}}

D .

\frac{7}{2 \sqrt{5}}

Answer: Option B
:
B

We have cos $θ$ =
$\frac{3}{5}$ and cos $ϕ$ =
$\frac{4}{5}$ .

Therefore $c o s (θ - ϕ) = c o s θ c o s ϕ + s i n θ s i n ϕ$

=
$\frac{3}{5}$ .
$\frac{4}{5}$ +
$\frac{4}{5}$ .
$\frac{3}{5}$ =
$\frac{24}{25}$

But $2 c o s^{2} (\frac{θ - ϕ}{2}) = 1 + c o s (θ - ϕ)$ = 1 +
$\frac{24}{25}$ =
$\frac{49}{50}$

∴ $c o s^{2} (\frac{θ - ϕ}{2})$ =
$\frac{49}{50}$ . Hence, $c o s (\frac{θ - ϕ}{2})$ =
$\frac{7}{5 \sqrt{2}}$ .

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Discussion Forum : Trigonometric Functions

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