Compound angle trig (1 Viewer)

[=)(=]

Q10b and d pls

 

[ExtremelyBoredUser]

You can simplify 10b and 10d to their sum and difference formulas.


Should be pre straightforward now. Construct triangles for each of the inverses, consider them alpha and beta respectively inside the sin(a+b) bracket and expand them how you would a regular sin(a+b) question, then construct the triangles for each of the separate functions and calculate respective values.

Here's an example of how it would like, for Q10b:

sin(sin^-1(4/5))cos(sin^-1(12/13) + cos(sin^-1(4/5)sin(sin^-1(12/13)

If you need more help, just reply and I can show the working out.

 

[=)(=]

You can simplify 10b and 10d to their sum and difference formulas.
View attachment 32579

Should be pre straightforward now. Construct triangles for each of the inverses, consider them alpha and beta respectively inside the sin(a+b) bracket and expand them how you would a regular sin(a+b) question, then construct the triangles for each of the separate functions and calculate respective values.

Here's an example of how it would like, for Q10b:

sin(sin^-1(4/5))cos(sin^-1(12/13) + cos(sin^-1(4/5)sin(sin^-1(12/13)

If you need more help, just reply and I can show the working out.

How do I simplify it from there

 

[Life'sHard]

How do I simplify it from there

The and should cancel like wise for

 

[ExtremelyBoredUser]

How do I simplify it from there

10B:

Simplified Version:
sin(sin^-1(4/5))cos(sin^-1(12/13) + cos(sin^-1(4/5)sin(sin^-1(12/13)

As lifehard mentioned, the sin and sin^-1 part would cancel out since if you drew the triangles and found the sin ratio for that triangle, you would get the exact value as the value inside the bracket. For the cos part, you would have to find the missing part and then do the cosine ratio.



cos(B) or cos(sin^-1(4/5)) = 3/5
cos(A) or cos(sin^-1(12/13) = 5/13

Hence;

4/5 * 5/13 + 3/5 * 12/13
= 56/65

 

[5uckerberg]

Okay while your question is answered I would like to introduce an alternative here

 

[=)(=]

Okay while your question is answered I would like to introduce an alternative here

thank you