Question

cos² 48⁰ - sin² 12⁰ = √5 + 1/8​

Answer 1

Qᴜᴇsᴛɪᴏɴ :-

Prove :- cos² 48⁰ - sin² 12⁰ = √5 + 1/8

Fᴏʀᴍᴜʟᴀ ᴜsᴇᴅ :-

• cos²A - sin²B = cos(A + B) * cos(A - B)
• cos36° = (√5 + 1)/4 .
• cos30° = (1/2)

Sᴏʟᴜᴛɪᴏɴ :-

Solving LHS,

→ cos²48° - sin²(12°)

Let A = 48° and B = 12° . Than, comparing it with cos²A - sin²B we get,

→ cos(48+12) * cos(48 - 12)

→ cos60° * cos36°

Putting Both values Now, we get ,

→ (1/2) * (√5+1)/4

→ (√5 + 1)/8 = RHS . (Proved).

Answer 2

$\huge \tt \underline \orange {Answer}$

$\bf \pink{ Given, }$

${\boxed{\huge \tt \red{Using \cos^2 48 \degree - \sin ^{2} 12 }}}$

$\tt \orange{Using : } \: \rm \purple{ \cos( A + B ) \: cos (A - B) = cos ^{2}A - sin ^{2} B}$

$\bf \green{ \implies cos(48 + 12) \: cos(48 - 12) }$

$\bf \green{ \implies cos(60) \: cos(36) }$

$\bf \green{ \implies \frac{1}{2} \: . \: cos(36) }$

$\huge \sf \pink{Now,}$

$\bf \orange{cos36 \degree = \frac{ \sqrt{5} + 1}{8} }$

$\bf \orange { \implies \frac{1}{2} \: . \: \frac{ \sqrt{5} + 1}{4} }$

$\bf \orange{ Therefore, \: proved \: that :- }$

$\longrightarrow \bf \green {cos ^{2} 48 - sin^{2} 12 \degree = \frac{ \sqrt{5} + 1}{8} }$