Mister Exam
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Identical expressions
- sin(-t/ four)=-sqrt(two)/ two
- sinus of ( minus t divide by 4) equally minus square root of (2) divide by 2
- sinus of ( minus t divide by four) equally minus square root of (two) divide by two
- sin(-t/4)=-√(2)/2
- sin-t/4=-sqrt2/2
- sin(-t divide by 4)=-sqrt(2) divide by 2
-
Similar expressions
- sin(-t/4)=+sqrt(2)/2
- sin(x)=-sqrt2/2
- cos*((pi(4x+48))/4)=-(sqrt(2)/2)
- sin(t/4)=-sqrt(2)/2
- cos(x)+sqrt((2-sqrt2)/2)*(sin(x)+1)=0
sin(-t/4)=-sqrt(2)/2 equation
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
___ /-t \ -\/ 2 sin|---| = ------- \ 4 / 2
$$\sin{\left(\frac{\left(-1\right) t}{4} \right)} = \frac{\left(-1\right) \sqrt{2}}{2}$$
Detail solution
Given the equation
$$\sin{\left(\frac{\left(-1\right) t}{4} \right)} = \frac{\left(-1\right) \sqrt{2}}{2}$$
- this is the simplest trigonometric equation
The equation is transformed to
$$\sin{\left(\frac{t}{4} \right)} = \frac{\sqrt{2}}{2}$$
This equation is transformed to
$$\frac{t}{4} = 2 \pi n + \operatorname{asin}{\left(\frac{\sqrt{2}}{2} \right)}$$
$$\frac{t}{4} = 2 \pi n - \operatorname{asin}{\left(\frac{\sqrt{2}}{2} \right)} + \pi$$
Or
$$\frac{t}{4} = 2 \pi n + \frac{\pi}{4}$$
$$\frac{t}{4} = 2 \pi n + \frac{3 \pi}{4}$$
, where n - is a integer
Divide both parts of the equation by
$$\frac{1}{4}$$
we get the answer:
$$t_{1} = 8 \pi n + \pi$$
$$t_{2} = 8 \pi n + 3 \pi$$
$$\sin{\left(\frac{\left(-1\right) t}{4} \right)} = \frac{\left(-1\right) \sqrt{2}}{2}$$
- this is the simplest trigonometric equation
Divide both parts of the equation by -1
The equation is transformed to
$$\sin{\left(\frac{t}{4} \right)} = \frac{\sqrt{2}}{2}$$
This equation is transformed to
$$\frac{t}{4} = 2 \pi n + \operatorname{asin}{\left(\frac{\sqrt{2}}{2} \right)}$$
$$\frac{t}{4} = 2 \pi n - \operatorname{asin}{\left(\frac{\sqrt{2}}{2} \right)} + \pi$$
Or
$$\frac{t}{4} = 2 \pi n + \frac{\pi}{4}$$
$$\frac{t}{4} = 2 \pi n + \frac{3 \pi}{4}$$
, where n - is a integer
Divide both parts of the equation by
$$\frac{1}{4}$$
we get the answer:
$$t_{1} = 8 \pi n + \pi$$
$$t_{2} = 8 \pi n + 3 \pi$$
Sum and product of roots
[src]
sum
pi + 3*pi
$$\pi + 3 \pi$$
=
4*pi
$$4 \pi$$
product
pi*3*pi
$$\pi 3 \pi$$
=
2 3*pi
$$3 \pi^{2}$$
3*pi^2
Numerical answer
[src]
t1 = 84.8230016469244
t2 = 5557.47740420034
t3 = 103.672557568463
t4 = -147.65485471872
t5 = -40.8407044966673
t6 = 78.5398163397448
t7 = -47.1238898038469
t8 = 12977.9192519794
t9 = 9.42477796076938
t10 = -91.106186954104
t11 = 34.5575191894877
t12 = 109.955742875643
t13 = -223.053078404875
t14 = -65.9734457253857
t15 = 3.14159265358979
t16 = -21.9911485751286
t17 = 28.2743338823081
t18 = 128.805298797182
t19 = -15.707963267949
t20 = 235.619449019234
t21 = -97.3893722612836
t22 = -72.2566310325652
t23 = 53.4070751110265
t24 = 59.6902604182061
t24 = 59.6902604182061