Question

Prove that $cos75°$ =$√3-1/2√2$​

Answer 1

Step-by-step explanation:

Given :-

Cos 75°

To find:-

Prove that Cos 75° = (√3-1)/(2√2)

Solution:-

Given that

LHS:-

Cos 75°

It can be written as Cos (45°+30°)

This is in the form of Cos(A+B)

Where A = 45° and B = 30°

We know that

Cos (A+B) = Cos A Cos B - Sin A sin B

On applying this formula to Cos (45°+30°)

=> Cos 45°×Cos 30° - Sin 45°× Sin 30°

=> (1/√2)×(√3/2) - (1/√2)×(1/2)

=> (1×√3)/(√2×2) - (1×1)/(√2×2)

=> √3/(2√2) - (1/2√2)

=> (√3-1)/2√2

=> RHS

LHS = RHS

Hence, Proved.

Answer:-

Cos 75° = (√3-1)/2√2

Used formulae:-

• Cos (A+B)=Cos A Cos B - Sin A sin B

• Sin 45° = 1/√2

• Cos 45° = 1/√2

• Sin 30° = 1/2

• Cos 30° = √3/2

Answer 2

Method A [Proof Without Words]

By the Pythagorean theorem,

$\overline{CA}^2=a^2+(\sqrt{3} a+2a)^2$

$\implies \overline{CA}^2=a^2+(3 +4\sqrt{3} +4)a^2$

$\implies \overline{CA}^2=(8+4\sqrt{3} )a^2$

$\implies \overline{CA}^2=2(4+2\sqrt{3} )a^2$

$\implies \boxed{\overline{CA}=\sqrt{2} (\sqrt{3} +1)a}$

Since $\cos75^{\circ}=\dfrac{\overline{CA}}{\overline{AB}}$,

$\cos75^{\circ}=\dfrac{a}{\sqrt{2} (\sqrt{3} +1)a}$

$\implies \cos75^{\circ}=\dfrac{1}{\sqrt{2} (\sqrt{3} +1)}$

$\implies \cos75^{\circ}=\dfrac{1}{\sqrt{2} (\sqrt{3} +1)}\times \dfrac{\sqrt{2} (\sqrt{3} -1)}{\sqrt{2} (\sqrt{3} -1)}$

$\implies \cos75^{\circ}=\dfrac{\sqrt{2} (\sqrt{3} -1)}{2(3-1)}$

$\implies \boxed{\cos75^{\circ}=\dfrac{\sqrt{6} -\sqrt{2} }{4}}$

Hence proven.

Method B[Cosine Additive Formula]

Formula: $\cos(A+B)=\cos A\cos B-\sin A\sin B$

$\implies \cos(30^{\circ}+45^{\circ})=\cos 30^{\circ}\cos 45^{\circ}-\sin 30^{\circ}\sin 45^{\circ}$

$\implies \cos75^{\circ}=\dfrac{\sqrt{3} }{2} \times \dfrac{\sqrt{2} }{2} -\dfrac{1}{2} \times \dfrac{\sqrt{2} }{2}$

$\implies \cos 75^{\circ}=\dfrac{\sqrt{6} }{4} -\dfrac{\sqrt{2} }{4}$

$\implies \boxed{\cos 75^{\circ}=\dfrac{\sqrt{6} -\sqrt{2} }{4}}$

Hence proven.