Trig Help!!?

Question

Find sin 2x if sinx = 1/5 and x is between 0 and 90 degrees

How would you go about solving this? The book says to draw a diagram, but i cant seem to be able to. I got as far as simplifying it to

sin2x = 2 sin1/5 cos1/5

Any help would be appreciated!!

Answer 1

sin x=1/5
Therefore cos x=sqrt(1-sin^2 x)
=sqrt(1-1/25)
=sqrt(24/25)
=2sqrt6/5 [x being in the first quadrant, only the positive value of cos x has been taken]
Now,
sin 2x
=2sin x cosx
=2*1/5*2sqrt6/5
=4sqrt6/25

Answer 2

Umm...

Sorry but your simplification is wrong:

sin2x = 2 sinx cosx, not 2 sin1/5 cos1/5

But you did start out correctly, rewriting sin2x into 2sinxcosx, this gives us an idea of what we need to do:

What you need to do is draw a right triangle in the first quadrant with a leg of 1 and a hypotenuse of 5. You get this from the definition of sine: opposite over hypotenuse.

Now solve for the other leg using pythagorean theorem:

5^2 = 1^2 + a^2
25 = 1 + a^2
24 = a^2
sqrt(24) = a
2sqrt(6) = a

Now you have all you need to solve for sin2x, back to it's rewritten form:

sin2x = 2 sinx cosx

You can now substitute your ratios:

2 (1/5) (2sqrt(6)/5) = 4sqrt(6)/25

Answer 3

Draw a right triangle that has one leg equal to one unit and the hypotenuse equal to five. By the Pythagorean theorem the other leg is
√24 = 2√6
So the cosine of x = (2√6)/5
Use the double angle formula:
sin2x = 2sinx cosx = 2(1/5)((2√6)/5) = (4√6)/25

Answer 4

Draw a triangle ABC with angle B = 90deg, and angle A = x.
Then from the definition of sine, you can make the length of BC = 1 and that of AC = 5.
Then from Pythagoras' Theorem:
AB = sqrt(5^2 - 1^2)
= sqrt(24)
= 2sqrt(6)
cos(x) = AB / AC = 2sqrt(6) / 5.
Now use sin(2x) = 2sin(x)cos(x):
sin(2x) = 2(1/5)(2sqrt(6) / 5)
= 4sqrt(6) / 25.

Answer 5

Remember that sin= opposite side / hypotenuse
So, opposite side to angle x is 1... and hypotenuse is 5.
If x is between 0 and 90 degrees then it must be located in the first quadrant of the cartesian.
To obtain the missing side (distance in x) you have to utilyze the formula c^2=a^2 + b^2. After that you'll discover that the sides of your triangle are
a=1
b= sqrt 24
c= 5 (hypotenuse)

Now, 2x is the double of the original angle x...hope it helps

Answer 6

Draw a right triangle whose hypotenuse is 5, its short leg is 1, and whose long leg is thus Sqrt(24)=2sqrt(6).
So sinx = 1/5 and cos x = 2sqrt(6)/5
Then sin 2x =2* 1/5 *2sqrt(6)/5 = 4sqrt(6)/25

Answer 7

sin(2x) = 2 sin(x) cos(x)
now sin(x) = 1/5 and sin^2(x) + cos^2(x) = 1 so 1/5^2 + cos^2(x) = 1
cos^2(x) = 24/25
cos x = 2/5 sqrt(6)
since we are in the first quadrant (0 < x < 90), both sin and cos are positiive.

sin(2x) = 2 sin(x) cos(x) = 2 (1/5) (2/5) sqrt(6) = 4/25 sqrt(6)

Answer 8

why not use inverse sin; you know: sin-1

sinx = 1/5 x = sin-1(1/5) then do sin[2sin-1(1/5)]
since 0 < x < 90, sinx must positive number less than 1

Answer 9

sin(2x) = 2sin(x)cos(x)

=2sin(x)(1-sin(x)^2)^(1/2)

=2(1/5)(24/5)^(1/2)

=sqrt(76/125)