Correct me

[Samin]

1.Evaluate the following using half angle identities
(a) tan (75◦)
(b) cos (120◦)
2. Evaluate cos (195◦) + cos (105◦) using sum-to-product/difference-to-product identities.

Answers

2. cos 195+cos 105=2cos(195+105)/2*(cos)195-95)/2
=2cos(300/2)*cos(90/2)
=2cos(150)*cos(45)
=2*-root 3/2* root 2/2
= -root 6/2=-1.224

1.(a) tan75=tan 150/2=1-cos 150/sin 150^2
=1-cos150/sin 150
=1+cos30/sin30
=1+root3/2(root just over 3)/1/2 (this whole thing is divided by 1/2)
=2+root 3

(b) cos 120=cos(240/2)=-whole root 1+cos 240 / 2(whole thing divided by 2) [CAN I GET DETAILED WORKINGS FOR THIS MATH?]
= -1/2

 

[Dr.Peterson]

You MUST learn to use parentheses! Rather than saying this, which is ambiguous despite all the words,

1+root3/2(root just over 3)/1/2 (this whole thing is divided by 1/2)

you should say this:

(1 + sqrt(3)/2)/(1/2)​

which is perfectly clear.

Similarly,

cos 195+cos 105=2cos(195+105)/2*(cos)195-95)/2

should be written as

cos(195) + cos(105) = 2 cos((195+105)/2)*(cos(195-95)/2)​

Do you see the difference?

I'll read your work after you make it readable.

Also, it may help if you tell us where you are unsure, so we can focus there.

 

[lex]

For 1(b) you seem to be using: [MATH]\qquad\cos^2{\left(\frac{x}{2}\right)=\frac{\cos(x)+1}{2}}\qquad[/MATH] with [MATH]x=240[/MATH]Then just use your angles now (and remember when you take the square root, that [MATH]\cos(120)[/MATH] is negative).

However I would have thought you would use the basic angles you know the answers for directly e.g. 60º

[MATH]\cos{x}=2\ cos^2\frac{x}{2}-1[/MATH] and let [MATH]x=120[/MATH].