Latest exam Aptitude Question SOLUTION: If A+B+C=90 degress, prove that sin2A+sin2B+sin2C = 4cosAcosBcosC

• Placement Papers
• /
• exam
• /
• Numerical Ability
• /
• Trigonometry

exam Question Papers:

(#M40078626) EXAM Numerical Ability question Keep an EYE

If A+B+C=90 degress, prove that
sin2A+sin2B+sin2C = 4cosAcosBcosC

Asked In exam (9 years ago)

Unsolved Read Solution (1)

Is this Puzzle helpful?   (12)   (6) Submit Your Solution Trigonometry

Read Solution (1) :

Leafy Snails  (9 years ago)

Since A + B + C = 90, sinA = cos(B+C), sinB = cos(A+C), sinC = cos(A+B)...

Sin2A + sin2B + sin2C
• 2sin(A+B)cos(A-B)+2sinCcosC
• 2sin(A+B)cos(A-B)+2cos(A+B)sin(A+B)
• 2sin(A+B) [cos(A-B) + cos(A+B)]
• 2sin(A+B) [cosAcosB + sinAsinB + cosAcosB – sinAsinB]
• 4sin(A+B)cosAcosB
• 4cosCcosAcoB
• 4cosAcosBcosC (proven)

Hope that helps XD

Is this solution Helpfull? Yes (1) | No (1)  

Please Login to Read Solution.