Im not even sure if this is 3u (1 Viewer)

[cameronS]

cos50+cos70 = cos 10

prove. Someone help? what topic is this?

 

[black_man]

could it be double angles? like cos (a + b) = cosacosb-sinasinb?

 

[currysauce]

you sure thats the answer?


I think this is from: cos S + cos T = 2cos 1/2(S+T) cos 1/2(S-T)

cos50 + cos70

= 2cos 1/2(120) cos 1/2(-20)
= 2cos 60 cos -10
= 2 x 1/2 x cos -10
= cos -10
= cos 10 (since i think cos(-a) = cos(a))


oh and obviously, this is the trigonometry topic.

 

[Slidey]

cos50+cos70 = cos 10

cos(60+10)=cos60cos10-sin60sin10
cos(60-10)=cos60cos10+sin60sin10

cos(60+10)+cos(60-10)=cos70+cos50
=2cos60cos10
=2*1/2*cos10
=cos10

 

[cameronS]

Thanks guys.

 

[cameronS]

bahh

how about sin80-sin20 = sin40

how do u know what values to use?

 

[Slidey]

sin80-sin20=sin40

Use the rule: sina-sinb=2cos([a+b]/2)sin([a-b]/2) (derived similar to the problem above)
cos60=1/2
RHS=2cos60sin40=sin40
If you can't do it by inspection, (a+b)/2=60, (a-b)/2 --> a+b=120, a-b=80, a=100, b=20
but 2cos60sin40=sin100-sin20=sin(180-100)-sin20
=sin80-sin20=LHS #

 

[~ ReNcH ~]

oh and obviously, this is the trigonometry topic.

My teacher went over "sums to products" and "products to sums" as part of integration...we only did it in MX2 - I don't know what topic it actually comes under in MX2 since it's not in the MX1 syllabus.

 

[Slidey]

Actually, it is in the ME1 syllabus. It's under: "The following results should not be taught."

 

[currysauce]

wait so we don't need to know product to sums and sums to products in 3unit?

its in the cambo book, so i did the exercise

 

[shafqat]

no its not in the course
pender included as an extension exercise, as it is useful for some 4u questions