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So I missed a week of school and the holidays started and to catch up I have been given some work and I was struggling to complete this as I haven't had any tuition on how to:
sin²X=0.25
Its been bugging me and I don't know what to do to get rid of the sin². Had it just been sin I would have 'inverse sined' a quarter
All help appreciated
Quote:
Originally Posted by acwilson96
Surely you realize that $\displaystyle \sin(x)=\pm 0.5~?$
No sorry i didnt, im doing higher maths(SQA) and im not sure that is in the syllabus or perhaps i just missed it. But what steps do I do to get rid of sin²?
If $\displaystyle u^2=0.25$ then $\displaystyle u=\pm 0.5~.$Quote:
Originally Posted by acwilson96
oh right yes, i understand that but i thought if you square root then it would be sin(root x )= 0.25
Quote:
Originally Posted by acwilson96
Well $\displaystyle \sin^2(x)=[\sin(x)]^2$.
Is the question really $\displaystyle \sin(\sqrt{x})=0.25~?$.
Which is it?
Ok i guess its just a rule you need to know that sin²x=(sinx)²
Il keep that in mind. So the answer would inverse sin of 1/2
Quote:
Originally Posted by acwilson96
$\displaystyle x=\pm\arcsin(0.5)$
