Question

1a. Let x be an arbitrary angle. Verify that (sin x + cos x)^2 = 1 + 2sinxcosx You may assume the Pythagorean identities for arbitrary angles.
b. Find an acute angle y such that sin y + cos y = squareroot (2).
c. Let x be an arbitrary acute angle. Prove that sin x + cos x >= 1.

Answer 1

shankvee need ur help!!!!!!!!!

Answer 2

a) (sin x+ cos x)^2= (sin x)^2 + (cos x)^2 + 2 sin x cos x
since (sin x)^2 + (cos x)^2 =1
so the answer is 1 + 2 sin x cos x
b) sin y+ cos y =sqrt(2)
square
1+2sin y cos y= 2
2sin y cos y=1
2sin y cos y =sin 2y
sin 2y=1
2y= 90(degrees)
y=45;
c)AM > GM
(sin x +cos x)/2 >= sqrt(sin x cos x)
sin x+ cos >sqrt(2sin 2x)
since greatest value of sin 2x =1
so sin x + cos x>sqrt(2)
so sin x +cos x>1

Answer 3

shankvee pls help me!!!!!!

Answer 4

tejeshwar help you in what?

Answer 5

Hello shankvee...I dont understand anything you type..

Answer 6

please can you explain to me some more...

Answer 7

are you there shankvee

Answer 8

ok be specific what did you not understand...
see (sin x)^2 + (cos x)^2 =1 since;
sin x=perpendicular/hypotenuse
cos x=base / hypotenuse now square and add since
(perpendicular)^2+(base)^2=(hypotenuse)^2

Answer 9

part b....you said square (1 + 2sin y cos y = 2)

Did you mean put it in Square root?

Answer 10

and part c you said AM > GM..what does that mean?

Answer 11

are you there sir/maam?

Answer 12

not sir i am also a student
Ok i'll make it simple for you just square the equation in part b and use the identity of part a then you get sin x cos x=1/2
so x=45

Answer 13

for the third i'll give you an easier answer see (cos x +sin x)^2 = 1+2sin x cos x that means (sin x +cos x)^2 is some value greater than 1 so sin x + cos x is also greater than 1

Answer 14

THANK YOU!

Answer 15

I have another problem! Can you help me with it?