alleryStart learning →In triangle ABC, cos((A + C)/2) equals
- cos(B/2)
- sin(B/2)
- −sin(B/2)
- sin B
Show answer and solution
Correct answer: B. sin(B/2)
(A + C)/2 = π/2 − B/2, so its cosine is sin(B/2).
MHT-CET Trigonometry - II — practice questions — page 2
Explore 40 free Trigonometry - II questions in MHT-CET Mathematics. Open each question or exercise to read its content, reference answers and available explanation.
Practise Trigonometry - II with Quill →Trigonometry - II questions and exercises with answers
Showing questions 31–40 of 40.
In triangle ABC, cos((A + C)/2) equalsRead the answer and worked solution →For the angles of a triangle, tan(A + B) reduces toRead the answer and worked solution →The triangle identity for squares of sines gives sin²A + sin²B + sin²C =Read the answer and worked solution →For the angles of any triangle, the sum sin2A + sin2B + sin2C isRead the answer and worked solution →Reducing 495° to a first-turn angle, sin 495° equalsRead the answer and worked solution →Removing two full turns, cos 930° reduces toRead the answer and worked solution →Using a period of 180°, tan 840° reduces toRead the answer and worked solution →In the second quadrant, sin(π − θ) is equal toRead the answer and worked solution →Which of these is the correct value of sin(3π/2 − θ)?Read the answer and worked solution →If sinθ = 1/2 with θ acute, what is the value of sin 3θ − 3 sinθ?Read the answer and worked solution →
Worked practice questions
For the angles of a triangle, tan(A + B) reduces to
- tan C
- cot C
- −cot C
- −tan C
Show answer and solution
Correct answer: D. −tan C
tan(A + B) = tan(π − C) = −tan C.
The triangle identity for squares of sines gives sin²A + sin²B + sin²C =
- 2 + 2 cosA cosB cosC
- 2 − 2 cosA cosB cosC
- 1 + 4 sinA sinB sinC
- 4 cosA cosB cosC
Show answer and solution
Correct answer: A. 2 + 2 cosA cosB cosC
The standard triangle identity is sin²A + sin²B + sin²C = 2 + 2cosA cosB cosC.
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