If 3 sin2 A + 2 sin2 B = 1 and 3 sin 2A − 2 sin 2B = 0, where A and B are acute angles then find the value of (A +2B)


3 years ago

Share

Answers : (1)

                                        

Dear Nirabhra, 


sin2B=3/2sin2A …(i)


And from the first equation,


3sin2A=1-2sin2B=cos2B …(ii)


cos (A + 2B) = cos A. cos 2B – sin A.sin 2B


3cosA.sin2A-32sinA.sin2A


3cosA.sin2A-3sin2AcosA =0


cos (A+2B)=0


∴ A+2B=π2,3π2 …(iii)


Given that < A < π/2 and 0 < B < π/2


⇒ 0<A+2B<π+π2


⇒ 0<A+2B<3π2 …(iv)


From (iii) and (iv),


A+2B=π2


 




Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.


All the best.


Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.


Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..


 


Askiitians Expert


Sagar Singh


B.Tech, IIT Delhi



3 years ago

Post Your Answer

More Questions On Trigonometry

Ask Experts

Have any Question? Ask Experts
Post Question
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!!
Click Here for details
sin10*sin30*sin50*sin70=
 
 
We have sinx*sin(60 – x)*sin(60 + x) = 1/4* sin3x sin10*sin30*sin50*sin70 =1/2*sin10*sin(60 – 10)*sin(60 + 10) =1/2*1/4*sin(3*10) =1/2*1/4*1/2 =1/16
 
Y RAJYALAKSHMI 19 days ago
COS(A-B)/COS(A+B)+COS(C+D)/COS(C-D)=0 THEN SHOW THAT tan A tan B tan C tan D=-1
 
 
STEP1:COS(A-B)/COS(A+B)=(COS(A)COS(B)+SIN(A)SIN(B))/(COS(A)COS(B)-SIN(A)SIN(B)) SIMILARLY OPEN THE EXPRESSION FOR THE OTHER TERM. STEP2:DIVIDE NUMERATOR AND DENOMINATOR IN BOTH CASES BY...
 
siddharth gupta one month ago
solve for x sin3x *sin(cube) x +cos3x *cos(cube) x=0 arun sir answered me but it was a tough method i need some easy way as i am in 11 th so a method which i couldget please answer>>>
 
 
Ans: It is a simple polynomial, you can easily solve. And if you have options, you can hit & trial.
  img
Jitender Singh one month ago
 
sir can u explain me how u solved the third last step not getting ?? is there any such identity??
 
rashmi one month ago
Let a,b,c E R and a (is not equal to 0) be such that (a+c) 2 2 ,then the quadratic equation ax 2 + bx + c = 0 has : Imaginary roots. Real roots. Exactly one real root lying in the...
 
 
Hello student, Please find the answer to your question below Givena,b,cER and a (is not equal to 0) Also given (a+c) 2 <b 2 So (a+c) 2 -b 2 <0 (a-b+c)(a+b+c)<0 So ax 2 +bx+c has...
  img
SHAIK AASIF AHAMED 11 days ago
the normal to the curve at P(x,y) meets the x axis at G. If the distance of G from the origin is twice the abscissa of P , then the curve is a) ellipse b) parabola c) ellipse or hyperbola
 
 
solution: for any curve the equation of normal at ( X1, Y1) (y- y1) = -1/m (x- X1) {where m = slope of tengent or say dy/dx at (x1, Y1) } normal meets the X axis at G.. so ( 0 - y1) = -1/m...
  img
Ajay Verma 5 months ago
 
but sir the answer says it can be both ellipse as well as hyperbola, so can u please prove this as an ellipse too.
 
vansh kharbanda 5 months ago
 
thank you
 
vansh kharbanda 5 months ago
see attachment and explain it
 
 
Ans: Hello Student, Please find answer to your question below Curve 1: Differentiate Slope of tangent: …......(1) Curve 2: Differentiate Slope of tangent: ...........(2) Since curve...
  img
Jitender Singh one month ago
View all Questions »