LockB's answer to LockB's Junior College 1 H2 Maths Singapore question.
It should be sqrt 65.
Anyway, outside the sin inverse there must be a multiplier 1/sqrt65 as part of the “reverse chain”.
Anyway, outside the sin inverse there must be a multiplier 1/sqrt65 as part of the “reverse chain”.
sorry, just to ask what is the reverse chain about? i never heard it before haha, i solved the qn based on my formula list....
also, is it possible to help take a look at the question i posted recently? i tried solving part a and got the answer wrong, i constantly got 1.84 (radian mode) after multiple tries and the ans key was 4.69
tried degree mode too but i got 4 instead of 4.69
also, is it possible to help take a look at the question i posted recently? i tried solving part a and got the answer wrong, i constantly got 1.84 (radian mode) after multiple tries and the ans key was 4.69
tried degree mode too but i got 4 instead of 4.69
Sorry, my mistake. I made a mistake with the format. I mixed it up with the arctan version [arctan means tan inverse]. There is no multiplier for arcsin [arcsin means sin inverse].
Your answer is correct.
The arctan version has a multiplier outside. Had it become arctan, the multiplier 1 / sqrt 65 is needed.
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Also, a second mistake of mine. The chain rule part doesn't apply here because your term in the bracket, x + 4, differentiates into 1 (so technically reversing the chain means dividing by 1).
However, if the term inside the bracket is like 2x + 3, things will be different.
Your answer is correct.
The arctan version has a multiplier outside. Had it become arctan, the multiplier 1 / sqrt 65 is needed.
--------------------------------------------
Also, a second mistake of mine. The chain rule part doesn't apply here because your term in the bracket, x + 4, differentiates into 1 (so technically reversing the chain means dividing by 1).
However, if the term inside the bracket is like 2x + 3, things will be different.
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