Math and Arithmetic

Algebra

Calculus

# Can sin equal 2?

###### Wiki User

###### 2009-12-17 00:14:34

No. Sin of any angle is always less than or equal to 1.

## Related Questions

###### Asked in Algebra, Trigonometry

### When are y equals sin x and y equals cos x times cos x equal?

y=sin x
y=cos x cos x
sin x = cos^2 x
sin x = 1-sin^2 x
sin x -1 + sin^2 x = 0
sin^2 x + sin x -1 = 0
Let y=sin x
y^2+y-1 = 0
This equation is of form ay^2+by+c=0
a = 1 b = 1 c = -1
y=[-b+/-sqrt(b^2-4ac)]/2a]
y=[-1 +/-sqrt(1^2-4(1)(-1)]/(2)(1)
discriminant is b^2-4ac =5
y=[-1 +√(5)] / 2
y=[-1 -√(5)] / 2
sin x = [-1 +√(5)] / 2
x = sin^-1 [-1 +√(5)] / 2] = 0.6662394 radians
x = sin^-1 [-1 -√(5)] / 2] = sin^-1 (-1.618) -- has no
solution
When x = 0.6662394 radians, sin x and cos x times cos x are
equal.

###### Asked in School Subjects, Math and Arithmetic, Algebra, Calculus

### How do you solve 2 sin squared x is equal to sin x?

There will be 4 possible solutions if all you are looking for is
the angles, so you will need to find out which quadrant your angle
is in.
2sin²x = sin x **Subtract sin x from both sides.
2sin²x - sin x = 0 **Then factor out sin x.
sin x(2sin x - 1) = 0 **Set each equal to zero. (AB=0 is the
same as A=0 OR B=0).
sin x = 0 or 2sin x - 1 = 0 to 2sin x = 1 to sin x =
1/2
At this point all that is left to do is find out where sin x = 0
or 1/2, which is 0, 180 for sin x = 0 or 30, 150 for sin x =
1/2.

###### Asked in Math and Arithmetic, Algebra, Calculus

### Is 2 cot x sin x cos x equals 2 - 2 sin 2 x an identity?

The easiest way to approach this problem is by rewriting the
left hand side entirely in terms of sin and cos and then
simplifying. To do so, use the fact that cot(x)=cos(x)/sin(x) to
get that
2*cot(x)*sin(x)*cos(x)=2*cos(x)/sin(x)*sin(x)*cos(x)=2*cos(x)²
Next, we will try to simplify the right hand side by factoring
and utilizing the formula cos(x)²+sin(x)²=1 which implies that
1-sin(x)²=cos(x)²
2-2sin(x)²=2*(1-sin(x)²)=2*cos(x)²
Since both sides can be simplified to equal the same thing, both
sides must always be equal, and the equation
2*cot(x)*sin(x)*cos(x)=2-2sin(x)² must be an identity

###### Asked in Geometry, Proofs

### How do you prove that two right triangles are congruent if the hypotenuse and an acute angle of one are equal respectively to the hypotenuse and acute angle of the other?

Let ABC and DEF be triangles which are right angled at A and D,
such that the hypotenuses BC and EF are equal and, without loss of
generality, angle B = angle E.Then
Then by the sine rule, BC/sin(A) = AC/sin(B) and EF/sin(D) =
DF/sin(E)
Since angle A = angle D = pi/2 radians, then sin(A) = sin(D) =
1
so that BC/sin(A) = BC while EF/sin(D) = EF
therefore, since the hypotenuses BC and EF are equal, the left
hand sides of the two equations are equal.
Therefore, AC/sin(B) = DF/sin(E)
then, since angle B = angle E, then sin(B) = sin(E) so that AC =
DF.
Also, angle C = pi/2 - angle B
and angle F = pi/2 - angle E
the right hand sides are equal so angle C = angle F.
Then in a manner similar to the above, we can show that AB =
DE.
Thus all three pairs of corresponding sides are equal and all
three pairs of corresponding angles are equal so that the two
triangles are congruent.