Question

cos square72-sins square54 evaluate this​

Answer 1

Answer:

-√5/4

Step-by-step explanation:

➡️ question given ➡️ cos²72°- sin²54°

➡️ question given ➡️ cos²72°- sin²54°➡️ cosø= sin(90-ø)

➡️ question given ➡️ cos²72°- sin²54°➡️ cosø= sin(90-ø) → cos 72°= sin(90-72)= sin18°

➡️ question given ➡️ cos²72°- sin²54°➡️ cosø= sin(90-ø) → cos 72°= sin(90-72)= sin18° → sin 18°=√5-1/4

➡️ question given ➡️ cos²72°- sin²54°➡️ cosø= sin(90-ø) → cos 72°= sin(90-72)= sin18° → sin 18°=√5-1/4➡️ sin ø=cos(90-ø)

➡️ question given ➡️ cos²72°- sin²54°➡️ cosø= sin(90-ø) → cos 72°= sin(90-72)= sin18° → sin 18°=√5-1/4➡️ sin ø=cos(90-ø) →sin54°= cos(90- 54)= cos36°

➡️ question given ➡️ cos²72°- sin²54°➡️ cosø= sin(90-ø) → cos 72°= sin(90-72)= sin18° → sin 18°=√5-1/4➡️ sin ø=cos(90-ø) →sin54°= cos(90- 54)= cos36° →cos36°=√5+1/4

➡️ question given ➡️ cos²72°- sin²54°➡️ cosø= sin(90-ø) → cos 72°= sin(90-72)= sin18° → sin 18°=√5-1/4➡️ sin ø=cos(90-ø) →sin54°= cos(90- 54)= cos36° →cos36°=√5+1/4➡️ cos²72°- sin 54°

➡️ question given ➡️ cos²72°- sin²54°➡️ cosø= sin(90-ø) → cos 72°= sin(90-72)= sin18° → sin 18°=√5-1/4➡️ sin ø=cos(90-ø) →sin54°= cos(90- 54)= cos36° →cos36°=√5+1/4➡️ cos²72°- sin 54°➡️ (√5-1/4)²- (√5+1/4)²

➡️ question given ➡️ cos²72°- sin²54°➡️ cosø= sin(90-ø) → cos 72°= sin(90-72)= sin18° → sin 18°=√5-1/4➡️ sin ø=cos(90-ø) →sin54°= cos(90- 54)= cos36° →cos36°=√5+1/4➡️ cos²72°- sin 54°➡️ (√5-1/4)²- (√5+1/4)²➡️(1/4)²[(√5-1)²-(√5+1)²]

➡️ question given ➡️ cos²72°- sin²54°➡️ cosø= sin(90-ø) → cos 72°= sin(90-72)= sin18° → sin 18°=√5-1/4➡️ sin ø=cos(90-ø) →sin54°= cos(90- 54)= cos36° →cos36°=√5+1/4➡️ cos²72°- sin 54°➡️ (√5-1/4)²- (√5+1/4)²➡️(1/4)²[(√5-1)²-(√5+1)²]➡️ 1/16[(√5)²+(1)²-2√5-(√5)²-(1)²-2√5]

➡️ 1/16×-4√5

➡️ 1/16×-4√5➡️-√5/4

hope it will be helpful