Question

Cos^4 A - Sin^4 A + 1 = 2cos A​

Answer 1

Answer:

Correct Question

$\mathsf{ {\cos(a) }^{4} - { \sin(a) }^{4} + 1 = 2 { \cos(a) }^{2}}$

Step-by-step explanation:

Given

• To prove

$\mathsf{ {\cos(a) }^{4} - { \sin(a) }^{4} + 1 = 2 { \cos(a) }^{2}}$

• Proof

Solving LHS

$\mathsf{ {\cos(a) }^{4} - { \sin(a) }^{4} + 1}$

$\mathsf{\implies {{(\cos(a) }^{2})}^{2} - { \sin(a) }^{4}+1}$

Now By equation

$\boxed{\bigstar \:{\sin{A}}^{2} - {\cos{A}}^{2} = 1}$

$\mathsf{\implies {{(\sin(a) }^{2} - 1 )}^{2} - {\sin(a)}^{4}+1}$

Now opening the brackets we have

$\mathsf{\implies {\sin(a) }^{4} + 1 -2 { \sin}^{2} (a) - { \sin(a) }^{4} + 1}$

$\mathsf{\implies 2-2{\sin(a)}^{2}}$

$\mathsf{\implies 2(1-{\sin(a)}^{2})}$

Now similarly by above equation we have

$\mathsf{\implies 2{\cos(a)}^{2}}$

LHS = RHS

Hence proved :)

Answer 2

Correct Question :---- cos⁴A - sin⁴A = 2cos²A

Formula to be used :---

• cos²A = 1 - sin²A
• (a-b)² = a² + b² - 2ab

Solving LHS now,

cos⁴A - sin⁴A + 1

→ (cos²A)² - sin⁴A + 1

→ (1-sin²A)² - sin⁴A + 1

using (a-b)² = a² + b² - 2ab now ,

→ 1 + sin⁴A - 2sin²A - sin⁴A + 1

→ 2 - 2sin²A

→ 2(1 - sin²A)

→ 2cos²A = RHS

Hence proved ...

(Hope it helps you)