Senior 1 Mathematics Trigonometric Identities The process of finding the answer

It's done, lz, wait, I'll take a picture with my phone and send it up.

Du Niang is too pumped, ** can't be passed on, let me briefly talk about the method.

In the first question, the first sin is induced by the formula, which becomes cos( 4- ) and then combines it with the two-angle difference sine formula, and finally becomes.

sin(4) = - root number 2 2

The second question cos95°=-sin5°, sin55°=cos35°, sin85°=cos5°, and then change the shape, it is easy to get the answer - root number 3 2

The third question is combined with the two-angle difference cosine formula, and the answer is cos

The fourth problem uses the formula to turn cos( - open, which is known to be the second quadrant angle, so cos = - root number 5 3, sin = root number 15 4

The answer is (2 root number 15 + root number 5) 12 (not necessarily right, you can do it yourself).

I don't want to talk about the fifth question.

According to this condition, the sixth problem is to move the right side of the inequality to the left side, and the deformation is easy to obtain cos(a+b) 0

Using the induction formula, it is -cosc 0, cosc 0

And because the inner angle of the triangle is in the open interval of 0 to , and cosc is less than 0, so the angle c is in the open interval of 2 to , and the angle c is an obtuse angle, so it is an obtuse triangle.

It's not easy to hit by hand, ask for it Finally, I wish LZ to learn and progress.

The fourth question is a direct formula, just bring it in.

Analysis: (1) Induction formula.

2) Sum of the difference, and the product of the difference.

3) Double angle formula, triple angle formula, half angle formula.

4) Magna formula.

There are mainly two angles and difference formulas, doubling angle formulas, power reduction formulas, and auxiliary angle formulas.

When x [0,1], note t = arcsinx, then t [0, 2], sint = x, then cost = 1-x ) gets: arcsinx = t = arccos (1-x).

When x [-1,0], arcsinx, then 2,0], sin =x, cos =1-x )

Thus cos( +cos =1-x ) 2, ] that is: +arccos (1-x) so arcsinx = arccos (1-x).

y=(1-cos2x) Sobasilica 2+sin2x 2 + 2(sin2x-cos2x) 2 + 5 2 2 2sin(2x- xiaotan4)+5 2 Suoyou with y [(5- 2) 2,(5+ 2) 2].

y=(finch cavity sin x) 2 + sinx cosx + 2-1 2cos2x+1 2sin2x+5 21 2(sin2x-cos2x)+5 2

Root number 2 2sin (2x-quarter of a sin shirt) +5 2 value range: root number state dust 2 2sin + 5 2

Use the inverse formula of the doubling angle formula first, and then use the one formula, and then it will be simple.