Help trigonometric questions 25, help trigonometric questions

3.Solution: tan(a+b)=(tana+tanb) (1-tanatanb).

tan∏/4=(tana+tanb)/(1-tanatanb)1=(tana+tanb)/(1-tanatanb)tana+tanb=1-tanatanb

1+tana)(1+tanb)

1+tana+tanb+tanatanb

1+1-tanatanb+tanatanb4.Solution: 1+(tana) 2=(seca) 2=1 (cosa) 2

i.e. 1+5 2=1 (cosa) 2

cosa=±1/√26

Again(sina) 2+(cosa) 2=1

sina=±5/√26

When A is the first or third quadrant angle, sina and cosa have the same name, and there is at this time.

sinacosa=5/26

When A is the second or fourth quadrant angle, Sina and Cosa have different names.

sinacosa=-5/26

Because cosa = 1 3, cos(a+b) = -3 5, and a and b are both acute angles.

So sina = 2 2 3, sin(a+b) = 4 5 substitution = (-3 + 8 2) 15

(1)..f(x)=(p/2)sin2wx-(1/2)cos2wx-1/2=(√1+p²)/2)sin(2wx-b)-1/2,tanb=1/p,2π/2w=π/2，w=2,√(1+p²)/2-1/2=1/2,p=1

f(x)=(2 2)sin(4x- 4)-1 2(2)cosine theorem a = b +c -2bccosa = bc finishing 1+2cosa = (b +c) bc 2cosa 1 2,0- 4<4a- 4 13 12

So 0

1. Because f(-x)=[x 1 3+x (-1 3)] 5=-f(x), f(x) is an odd function.

Let the annihilation be defeated x 1 3=t, the original formula is (t-t one-part) divided by 5, x belongs to 0 to positive infinity, t belongs to 0 to 1, (t-t one-part) is the subtraction function, and f(x) is the subtraction function. Oak sails.

In the same way, x belongs to negative infinity, and to 0 is f(x) is an increasing function.

2. Algebraically f(4)-5f(2)g(2) and f(9)-5f(3)g(3) both = 0, so f(x 2)-5f(x)g(x) is equal to 0

The sufficient and necessary conditions for vector parallelism are x1y2-x2y1=0. So according to this condition, the equation can be listed: sina (3sina-2) - (1-4cos2a) = 0 (alpha cannot be played, a is used instead) to solve sin squared a-2sina = 0, and at the same time divided by cosa, the solution is tan squared a-2tana = 0, tana = 2 or 0, because of the alpha value range, tana can not be equal to 0

Then we find the final value of -1 in the equation of tana = 2 generations

Because a b, sin is multiplied by (3sin -2)—1 times (1-4cos2) = 0, (note that cos2 = 1-2 sin squared). Simplification yields: 5sin squared + 2sin -3=0 to find sin = 3/5 or 1 and because 0< <2, 1 is rounded.

So tan = 3/4Using the tangent of the difference between the two angles, we can get the value of your solution to be minus 1/7

Vector a b, sina 1=(1-4cos2a) (3sina-2), after solving sina = 3 5, and then find cosa=4 5, tana=3 4, and substitute tana=3 4 to find the result.

The root number of minus half is three, a = 150 degrees.

Analysis: from a and b are the inner angles of the triangle, sina and sinb are both greater than 0, and then c is determined to be an obtuse angle, using the induction formula and the inner angle of the triangle and the theorem to simplify the left side of the known equation, sinb=-2sinacosc, and then by sinb=sin(a+c), the sine function formula of the sum and difference of the two angles is used to simplify, and then the basic relationship between the trigonometric functions of the same angle is simplified, and tanc=-3tana is obtained. The maximum value of tanb can be obtained by using the induction formula and the inner angle sum theorem of the triangle to -tan(a+c), and the tangent function of the sum and difference of the two angles is used to simplify it, and the range of tanb is obtained by using the basic inequality after deformation

Answer: Solution: sina 0, sinb 0, sinb sina = 2cos(a+b)=-2cosc 0, i.e. cosc 0, c is an obtuse angle, sinb = -2sinacosc, and sinb = sin(a+c) = sinacosc+cosasinc, sinacosc+cosasinc=-2sinacosc, i.e. cosasinc=-3sinacosc, tanc=- 3tana，tanb=-tan（a+c）=-(tana+tanc 1-tanatanc) =-(-2tana 1+3tan2a) =2 (1 tana+3tana)

2 2 root number 3 = (root number 3) 3 if and only if.

1 tana=3tana, that is, take the equal sign when tana=(root number 3) 3, then the maximum value of tanb is (root number 3) 3

Using a +1-1 will solve the problem. Try it.

From (3 4) cosa>1 we get (3 4) cosa) 0, so cosa<0, then the terminal edge of a falls in the second, third quadrant (cosine is negative).

Hope it helps.

Because 3 4 is less than 1, cosa<0, so the terminal side of a is in the second and fourth quadrants.