A few trigonometric function problems, everyone can help

2 Solution: You may wish to set ag=1

Obviously mag= nag= 6, agn= - and amg=5 6- ,ang= - 6 from the sinusoidal theorem, am sin =ag sin(5 6- )an sin( -=ag sin( - 6) gives am=sin sin(5 6- )an=sin sin( - 6).

s1=1/2×ag×am=sinα/2sin(5π/6-α)s2=1/2×ag×an=sinα/2sin(α-/6)y=1／s1²+1／s2²

4[sin²(5π/6-α)sin²(α/6)]/sin²α2[1-cos(5π/3-2α)+1-cos(2α-π/3)]/sin²α

2×/sin²α

2[2-2cos(2/3π)cos(π-2α)]/sin²α2(2-cos2α)/sin²α

2(1-cos2α)/sin²α+2/sin²α4+2/sin²α

Known by 3 2 3.

So when = 2, y has a minimum value of 6

When = 3 (n coincides with c) or 2 3 (m coincides with b), y has a maximum value of 20 3

3 Solution: From the sinusoidal theorem, s abc = 1 2ab acsina = 3a is obtained by s ade = s quadrilateral decb.

s ade = 1 2s abc = 3 2a and again. s ade=1 2ad aesina= 3x 4 ae is derived from: ae=2a x

There is by the cosine theorem.

cosa=(ad +ae -de) (2 ad ae)=1 2 Substituting ad=x, de=y, and .

y²=x²+4a4/x²-a²

It is easy to get from the above equation that if and only if x = 2a, y has a minimum value of 3a

3 Solution: From the sinusoidal theorem, s abc = 1 2ab acsina = 3a is obtained by s ade = s quadrilateral decb.

s ade = 1 2s abc = 3 2a and again. s ade=1 2ad aesina= 3x 4 ae is derived from: ae=2a x

There is by the cosine theorem.

cosa=(ad +ae -de) (2 ad ae)=1 2 Substituting ad=x, de=y, and .

y²=x²+4a4/x²-a²

It is easy to get from the above equation that if and only if x = 2a, y has a minimum value of 3a

Solution: From point D, the perpendicular line of AC and Xiaomu Xun BC is made respectively, and the perpendicular balance is E and F respectively, so Ac=AE+EC=AE+DF=B*Sinb+A*Cosa;

easy！1) Y has a maximum value when x+6/6 = 2k +2/2.

At this point, x = 2k + 3 parts

Refers to the game 2) which teases the value range of Li Jing -2 to 2

The value range [-2,2] -1 takes the maximum value: x+ sock imitation 6=k + closed fiber2 push-out state calendar x=k + 3

1) When the function Yu Tangerine sells y to obtain the maximum value of Brother Wu, x+6=2n + 2 x=2n+3

Since c=90° and da=2dc, according to the theorem: the right-angled side corresponding to 30° in a right-angled triangle is half the length of the hypotenuse, knowing that dac=30°

Because da is the angular bisector of bac, bad=dac=30° so bac=30°+30°=60°

So b=90°-60°=30°

So tanb = 3 3

Well, that's what you did.

tandac = 2 root number 15 6 root number 5 = root number 3 3 DAC is an acute angle.

Then DAC = 30°

AD is the angular bisector.

Then bad=dac=30°

c = 90°, then b = 90-60 = 30°

tanb = root number 3 3

(Written on the mobile phone, it is inconvenient to read, forgive me).

tan because tanbad=tan dac, so tan bac (tan because the angle c is 90 degrees, so tanb tan under the root

Answer: 3 7sinx + 4 7cosx

5 7 (3 nuclei touch 5sinx+4 5cosx), so that cosfai=3 changes to 5, sinfai=4 5, then the original formula = 5 7(cosfaisinx+sinfaicosx).

5/7sin(x+fai）.

According to the auxiliary angle formula, go to the book to see the specific formula.

Just keep it simple. cos2x=1-2sin²x

Analysis: First, where it is not strict: sint=3 5 x=37 180 should be:

t=arcsin(3 5)ps:whether t is accurately taken may affect the final result two, where it is not rigorous: let x=sint, forget to specify the value range of t three, the misconception a should first use the commutation method to find the indefinite integral, and then substitute the value to find the definite integral b instead of both at the same time ps:

It's not that A is wrong, it's that B is too prone to error in the process of solving problems.

Use the special value method: m = -20 degrees, period = 200 degrees, quarter period is equal to 50 degrees, n = -20 + 50 = 30 degrees, the minimum point is 130 degrees, and the answer is b