Senior 1 Mathematics. Solve the detailed process without skipping steps.

Solution: Question 1:

From the title, the original formula (sin cos) sin cos (the terms of the numerator and the formula are divided by the denominator) sin cos cos sin tan 1 tan 2 1 2 5 2

So, choose C.

Question 2: Because sin cos 1, so 1 2sin cos sin cos 2sin cos (using the formula of perfect sum of squares) (sin cos).

So the root number (1 2sin cos) the absolute value (sin cos) So, choose d.

Question 3: From the question, known cos -3 5,

By the formula sin cos 1, substituting the formula into sin 4 5, and because (3 2), then sin < 0, so sin -4 5, then tan sin cos 4 3.

ps: I don't have time to search for the name of the corner, but it should be no problem to understand, and I'll just look at it.

d1=sin^2(a)+cos^2(a),1+2sin(a)cos(a）=sin^2(a)+cos^2(a)+2sin(a)cos(a)=[sin(a)+cos(a)]^2

The answer is |sin(a)+cos(a)|Sin 2(a) represents the square of sin(a).

。Hope.. Pick.. Accept...

f'(x)=4x-1 x=(4x2-1) xlet f'(x)=0, note that the original function defines the domain as x>0, and we find x=1 2 when x (0,1 2), f'(x)<0, f(x) monotonically decreasing;

When x (1 2, + f'(x) >0, f(x) monotonically increasing.

Because [k-1,k+1] is in the defined domain, k-1>0, k>1;

Because [k-1,k+1] is not a monotonic interval, k-1<1 2, so k<3 2

In summary, k (1,3 2).

The solution is known by the sinusoidal theorem a sina=b sinb.

Then 2asinb = 3b

Becomes 2sinasinb = 3sinb

Because sinb 0

Then 2sina = 3

Knowing sina = 3 2

and a is an acute angle.

then a = 3, so choose a

Answer: A According to the sine theorem there are:

a sina=b sinb=c sinc=2r, substitution of the conditional formula 2asinb= 3b, there is:

2sinasinb=√3sinb>0

So: sina = 3 2

Solution: a=60° (obtuse angle does not conform to rounding) choose a

Since 2asinb = 3b, then sinb = 3b 2a, by the sine theorem, b sinb = a sina gives sinb = bsina a, so bsina a = 3b 2a, so sina = 3 2, and because the triangle is an acute triangle, a = 60°

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