Question

Cos 60° • Cos 42° • Cos 66° • Cos 78° = $\frac{1}{16}$​

Answer 1

Step-by-step explanation:

Given :

\bullet\ \; \sf \dfrac{\sqrt{a^2}-\sqrt{x^2}}{\sqrt{a^2}+\sqrt{x^2}}∙

a

2

+

x

2

a

2

−

x

2

To Find :

Derivative of the given value

Solution :

Let ,

\begin{gathered}\sf y=\dfrac{\sqrt{a^2}-\sqrt{x^2}}{\sqrt{a^2}+\sqrt{x^2}}\\\\\to \sf y=\dfrac{a-x}{a+x}\end{gathered}

y=

a

2

+

x

2

a

2

−

x

2

→y=

a+x

a−x

Differentiate with respect to x on both sides ,

\to \sf \dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(\dfrac{a-x}{a+x}\bigg)→

dx

dy

=

dx

d

(

a+x

a−x

)

d/dx(u/v) = (vu' - uv') / v²

u' ➠ du/dx

v' ➠ dv/dx

\to \sf \dfrac{dy}{dx}=\dfrac{(a+x)\frac{d}{dx}(a-x)-(a-x)\frac{d}{dx}(a+x)}{(a+x)^2}→

dx

dy

=

(a+x)

2

(a+x)

dx

d

(a−x)−(a−x)

dx

d

(a+x)

\to \sf \dfrac{dy}{dx}=\dfrac{(a+x)(-1)-(a-x)(1)}{(a+x)^2}→

dx

dy

=

(a+x)

2

(a+x)(−1)−(a−x)(1)

\to \sf \dfrac{dy}{dx}=\dfrac{-a-x-a+x}{(a+x)^2}→

dx

dy

=

(a+x)

2

−a−x−a+x

\leadsto \sf \pink{\dfrac{dy}{dx}=\dfrac{-2a}{(a+x)^2}}\ \; \bigstar⇝

dx

dy

=

(a+x)

2

−2a

★

Answer 2

Solution:

Hello Mate!!!

I guess your question will be Cos 6° • Cos 42° • Cos 66° • Cos 78° = 1/16, not Cos 60° • Cos 42° • Cos 66° • Cos 78° = 1/16.

To Prove - Cos 6° • Cos 42° • Cos 66° • Cos 78° = $\frac{1}{16}$

Proof -

=> Cos 6° • Cos 42° • Cos 66° • Cos 78° = $\frac{1}{16}$

=> (Cos 6° • Cos 66°) • (Cos 42° • Cos 78°)= $\frac{1}{16}$

Multiplying and dividing by 4,

=> $\frac{1}{4}$ • (2 • Cos 6° • Cos 66°) • (2 • Cos 42° • Cos 78°) = $\frac{1}{16}$

We know that,

cos C + cos D = 2•cos((C+D)/2)•cos((C-D)/2)

So, using inverse relation of it,

=> $\frac{1}{4}$ • (Cos 60° + Cos 72°) • (Cos 120° + Cos 36°) = $\frac{1}{16}$

=> $\frac{1}{4}$ • ($\frac{1}{2}$ + Cos 72°) • ($\frac{-1}{2}$ + Cos 36°) = $\frac{1}{16}$

(Cos 72° = $\frac {√5-1}{4}$ & Cos 36° = $\frac {√5+1}{4}$ )

=> $\frac{1}{4}$ • ($\frac{2}{4}$ + $\frac {√5-1}{4}$ ) • ($\frac{-2}{4}$ + $\frac {√5+1}{4}$ ) = $\frac{1}{16}$

=> $\frac{1}{4}$ • ($\frac{√5+1}{4}$) • ($\frac{√5-1}{4}$) = $\frac{1}{16}$

=> $\frac{(√5-1)(√5+1)}{64}$ = $\frac{1}{16}$

=> $\frac{( (√5)^2 - (1)^2 )}{64}$ = $\frac{1}{16}$

=> $\frac{( 5 - 1 )}{64}$ = $\frac{1}{16}$

=> $\frac{(4)}{64}$ = $\frac{1}{16}$

=> $\frac{1}{16}$ = $\frac{1}{16}$

HENCE PROVED.!!!!

HOPE IT HELPS!!!!

PLS MARK IT BRAINLIEST AND FOLLOW ME BCOZ IT TOOK A LOT OF TIME!!!!