C3 Trig help required

sorry i was lazy and didnt look at your method but i did it by using the fact that
cos=sqrt(1-sin^2).

now if you sub in arcsin(0.5) to sin^2 your left with 0.5^2 which is 0.25.

So sin(x)=sqrt(1-0.25)
sin(x)=sqrt(3)/2

this gives x values of 60 and 300

hope that helps. using that first identity is a nice little trick

I'm stuck on q5a and here's my working out

Let A=arcsin1/2
SinA=1/2
A=30°
A=180-30=150°

When A=30°
Sinx=cos30°
Sinx=√3 /2
x=60° <-- solution

When A=150°
Sinx=cos150°
Sinx= -√3 /2

From that ^ messy diagram (sorry),
x=60° <-- solution

But the textbook answer is x=60, 300°. I'm pretty sure I went wrong in the final step, but I can't see how you get x=300° using special triangles in the final step and need someone to help me out here please. I know that I can use a calculator to get x=300°, but I want to do it without a calculator.

Thanks so much!

sorry i was lazy and didnt look at your method but i did it by using the fact that
cos=sqrt(1-sin^2).

now if you sub in arcsin(0.5) to sin^2 your left with 0.5^2 which is 0.25.

So sin(x)=sqrt(1-0.25)
sin(x)=sqrt(3)/2

this gives x values of 60 and 300

hope that helps. using that first identity is a nice little trick

arcsin(1/2) has a numerical value which you can find on the calculator (or should know by heart).

So you can replace arcsin(1/2) by this value.

Can you see where to go now?

cos=sqrt(1-sin^2) I understand this identity now.

But don't you mean cos=√0.75 ? That's what the original formulae you stated implies.

EDIT: ok sorry i didnt properly answer your question. firstly you've drawn the angle in the wrong place in your diagram, it should be at the origin. also when your using that diagram you have to use CAST. when the angle is in the top-left area where you've been drawing sin is positive so you wouldnt get -srqt3/2.

How am I supposed to even know where the angle is? Sinx=-√3 /2 so sin is - and sin is only - in quadrants 3 and 4. So is the triangle in quadrant 3 or 4?

Thanks

tbh i'm not really sure why its not both...the same applies to the + too, so i feel like the answers should be 60, 120, 240, 300..are you sure about it just being 60 and 300?

No more solutions

Book's wrong IMHO, values are 60, 120, 240, 300 as Tedward said.

How?

IF sin x = +/- rt(3)/2 then the result follows x = 60 or 120 for the positive value, and 240 or 300 for the negative value.

IF sin x = +/- rt(3)/2 then the result follows x = 60 or 120 for the positive value, and 240 or 300 for the negative value.

When sinx=√3/2

Using special triangles, x=60°
Using CAST diagram, x=180-60, so x=120°

When sinx= -√3 /2

For sinx=√3 /2 we said x=60°, so when sinx=-√3 /2 x=-60 (not a solution in range)
x=360-60
x=300°
Using CAST diagram, x=180+60=240°

Right?

X

Can you help me in q5b and in q5c please?

In 5b I got x=45, 135, 225, 315°. But my textbook say only x=45, 225°. What did you get?

Similar thing in q5c. I got x=0, 60, 120, 180, 240, 300, 360°. But my textbook says only x=0, 360°. What did you get?

Thanks a lot for all the help, much appreciated!

I presume your answers are different to the book.

One thing I'd like to check is what are your definitions of arctan, arccos, arcsin? Do they stipulate a range. I'd gone along with the idea intially that arcsin(1/2) could take multiple values, but strictly speaking as a function, it can only take one value. So, what are the definitions?