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
92 views
in Trigonometry by (114k points)
closed by
The value of sin2 450 - cos2 150 is
1. \(-\frac{\sqrt{3}}{4}\)
2. \(\frac{1}{4}\)
3. \(\frac{-1}{2}\)
4. \(\frac{2}{3}\)

1 Answer

0 votes
by (113k points)
selected by
 
Best answer
Correct Answer - Option 1 : \(-\frac{\sqrt{3}}{4}\)

Concept:

  • \(\rm 2 cos^{2} \theta= 1+ cos2\theta\)  
  • \(\rm 2 sin^{2} \theta= 1- cos2\theta\)  

 

Caculation:

We know that , sin 45o = \(\rm \frac{1}{\sqrt{2}}\) , 

∴ sin2 45o = \(\rm \frac{1}{{2}}\)  

\(\rm \cos^{2} \theta= \frac {(1+ \cos 2\theta)}{2}\)

⇒ cos2 150 =  \(\rm \frac{(1+\cos 2\times 15)}{2}\) =  \(\rm \frac{(1+ \cos 30)}{2}\) = \(\frac{2+\sqrt{3}}{4}\)

⇒ cos150 = \(\frac{2+\sqrt{3}}{4}\)

Now,

sin2 450 - cos2 150  = \(\rm \frac{1}{{2}}\)-  \(\left ( \frac{2+\sqrt{3}}{4}\right )\)

⇒sin2 450 - cos2 150  = \(-\frac{\sqrt{3}}{4}\)  

The correct option is 1. 

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

...