Question

express as a single fraction the exact value of sin75

Answer 1

\[\huge \sin75^0 = \sin (45^0+ 30^0)\]
\[\huge= \sin 45^0 \cos 30^0 + \cos45^0 \sin 30^0\]
\[\huge= \frac{1}{ \sqrt2} \frac{\sqrt3}{2} + \frac{1}{ \sqrt2} \frac{1}{2} \]
\[\huge= \frac{\sqrt3}{2\sqrt2} + \frac{1}{ 2\sqrt2} = \frac{\sqrt3 +1}{2\sqrt2}\] is the required answer.

Answer 2

@awsome500

Answer 3

Since 75= 45+30

Answer 4

so i used 45 + 30 in place of 75 and also we don't know the direct value of 75 degrees so we need to split 75 as 45+ 30

Answer 5

@awsome500 any further query

Answer 6

thank you

Answer 7

@awsome500 Medal please...

Answer 8

cos 45 = \[\frac{ \sqrt{2} }{ 2}\]

Answer 9

sin75 will the answer be \[\frac{ \sqrt{6}+\sqrt{2} }{ 4 }\] @mathmale

Answer 10

Without going through every detail of the other person's attempts to show you the way, I'll say his approach does look very appropriate. Why are you in doubt over his answer? I'd suggest you review his work and ask questions about anything not clear to you. I'd be delighted to respond in that case. I'd rather not be asked to go through someone else's entire solution of a problem and judge whether it's correct or not.

Answer 11

@awsome500 both of those answers are equivalent; the one you propose is simply the one @dpasingh gave after being rationalized.

Answer 12

thank you i understand

Answer 13

\[\frac{1+\sqrt{3}}{2\sqrt{2}} = \frac{1+\sqrt{3}}{2\sqrt{2}} *\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}(1+\sqrt{3})}{2\sqrt{2}\sqrt{2}} = \frac{\sqrt{2}+\sqrt{6}}{4}\]

Answer 14

Unless the problem specifically requests that you rationalize the denominator, either one should be acceptable.

Answer 15

@mathmale please help me on one more question express \[\frac{ \cot \chi \sin \chi }{ \sec }\] as a single trigonometric function in simplest form for all values of x for which it is defined

Answer 16

Both the secant and the cotangent can be re-written in terms of sine and cosine. Would you please do that now, before considering any other action?

Answer 17

okay cot will be cos over sin
and sec will be 1 over cos

Answer 18

Good! Now let's put that all together. Here's what we now have:\[\frac{ \cos x }{ \sin x}\frac{ \sin x }{ 1 } ~divided~by~\frac{ 1 }{ \cos x }\]

Answer 19

Try writing that out on paper. Cancel wherever possible.

Answer 20

so we cancel sin which will leave us with cos over 1 divided by 1over cos

Answer 21

doesnt cos get canceled out to

Answer 22

I'd like to see what you have done; then I'd be in a better position to answer that. Can you upload an image? If not, type out what you have done.

Answer 23

Yeah, so you canceled sin(x) and now you have

\(\Huge\color{blue}{ \cal \frac{Cos(x)}{1}=\frac{Cos(x)}{1} }\)

well, you reached the point a=a .

Answer 24

but sec = 1over cos @SolomonZelman

Answer 25

I didn't see the actual thing, you drew cos x /1 on the right side.. OK.

So you have

\(\Huge\color{blue}{ \cal \frac{Cos(x)}{1}=\frac{1}{Cos(x)} }\)

you would cross multiply. Tell me what you get ?

Answer 26

Or, you can just use a common sense, knowing that Cos(x) can't equal anything else, other than 1.

Answer 27

If you were proving/verifying, then both sides aren't equal. because they are not equal for all values of x.

Answer 28

did u flip the recpical of sec which was 1 over cos and it became cos over 1