Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
106 views
in Trigonometry by (20 points)
edited by

(a) Write the product 2 sin⁡(2π/9) cos⁡(π/9) as a sum of function values.

(b) Express cos⁡55°cos⁡45° as a sum.

(c) Express the product sin⁡6θ cos⁡4θ as a sum.

(d) Write the sum sin⁡(π/9)+sin⁡(2π/9) as a product of function values.

(e) Express sin⁡55°+sin⁡45° as a product.

(f) Use the product-to-sum to evaluate cos⁡3π/2 cos⁡π/2.

(g) Use the sum-to-product to evaluate cos⁡3π/2+cos⁡π/2.

Please log in or register to answer this question.

1 Answer

0 votes
by (37.8k points)

(a) 2 sin \(\frac {2\pi}{9}\) cos \(\frac {\pi}{9}\) 

=  sin (\(\frac {2\pi}{9}\) + \(\frac {\pi}{9}\)) + sin (\(\frac {2\pi}{9}\) - \(\frac {\pi}{9}\))

= sin (\(\frac {3\pi}{9}\)) + sin (\(\frac {\pi}{9}\))

= sin (\(\frac {\pi}{9}\)) + sin (\(\frac {\pi}{9}\))

(b) cos 55° cos 45° 

\(\frac 12 \) (cos (55° + 45°) + cos (55° - 45°))

\(\frac 12 \) (cos 100° + cos 10°)

(c) sin 6 θ cos 4 θ 

\(\frac 12 \) (sin 6 θ + 4 θ) + sin ( 6 θ - 4 θ)

\(\frac 12 \) (sin 10 θ + 2 θ) 

(d) sin \(\frac {\pi}{9}\) + sin \(\frac {2\pi}{9}\) 

= 2 sin \((\frac {\frac {\pi}{9} + \frac {2\pi}{9}}{2}) cos (\frac {\frac \pi9 -\frac {2\pi}{9}}{2})\) 

= = 2 sin \((\frac {3\pi}{18}) cos ( \frac {-\pi}{18})\)  

(e) sin 55° + sin 45°  

= 2 sin \((\frac {55°+45°}{2}) cos (\frac {55°-45°}{2})\) 

= 2 sin \((\frac {100°}{2}) cos (\frac {10°}{2})\) 

= 2 sin 50° cos 5°

(f) cos \(\frac {3\pi}{2} cos \frac {\pi}{2} \) =  0 x 0 

= 0 (∵ \(cos \frac {\pi}{2} = \) 0 & cos \(\frac {3\pi}{2}\) = 0)

(g) cos \(\frac {3\pi}{2} + cos \frac {\pi}{2} \) 

= 0 + 0 = 0

or

(f) cos \(\frac {3\pi}{2} cos \frac {\pi}{2} \) 

\(\frac 12 \) ( cos ( \(\frac {3\pi}{2} cos \frac {\pi}{2} \)) + cos ( \(\frac {3\pi}{2} - \frac {\pi}{2} \) ))

\(\frac 12 \)  ( cos ( \((cos 2\pi + cos \pi) = \) \(\frac 12 \) (1 -1) = 0

(g) cos \(\frac {3\pi}{2} + cos \frac {\pi}{2} \) 

= 2 cos \(\frac {\frac {3\pi}{2}+\frac {\pi}{2}}{2} cos \frac {\frac {3\pi}{2}-\frac {\pi}{2}}{2} \) 

= 2 cos π  cos π/2

= 2 x -1 x 0 = 0

Related questions

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...