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3 years ago

If sin∅+cos∅ = 1 , find sin∅.cos∅.

Mathematics

2 answers:

Katena32 [7]3 years ago

7 0

<h3>Answer: 0</h3>

=============================================

Explanation:

The original equation is in the form a+b = 1, where

a = sin(theta)

b = cos(theta)

Square both sides of a+b = 1 to get

(a+b)^2 = 1^2

a^2+2ab+b^2 = 1

(a^2+b^2)+2ab = 1

From here notice that a^2+b^2 is sin^2+cos^2 = 1, which is the pythagorean trig identity. So we go from (a^2+b^2)+2ab = 1 to 1+2ab = 1 to 2ab = 0 to ab = 0

Therefore,

sin(theta)*cos(theta) = 0

Ganezh [65]3 years ago

4 0

Answer:

sin ∅ cos ∅ = 0.

Step-by-step explanation:

(sin∅+cos∅)^2 = 1^2 = 1

(sin∅+cos∅)^2 = sin^2∅ + cos^2∅ + 2sin ∅ cos ∅ = 1

But sin^2∅ + cos^2∅ = 1, so:

2sin ∅ cos ∅ + 1 = 1

2 sin ∅ cos ∅ = 1 - 1 = 0

sin ∅ cos ∅ = 0.

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find x if the distance between points L and M is 15 and point M is located in the first quadrant. L=(-6,2) M=(x,2)

aivan3 [116]

Answer:

Therefore,

$x = 9$

Step-by-step explanation:

Given:

Let,

point L( x₁ , y₁) ≡ ( -6 , 2)

point M( x₂ , y₂ )≡ (x , 2)

l(AB) = 15 units (distance between points L and M)

To Find:

x = ?

Solution:

Distance formula between Two points is given as

$l(LM)^{2} = (x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$

Substituting the values we get

$15^{2}=(x--6)^{2}+(2-2)^{2}\\\\225=(x+6)^{2}$

Square Rooting we get

$(x+6)=\pm \sqrt{225}=\pm 15\\\\x+6 = 15\ or\ x+6 = -15\\\\x= 9\ or\ x = -21$

As point M is located in the first quadrant

x coordinate and y coordinate are positive

So x = -21 DISCARDED

Hence,

$x = 9$

Therefore,

$x = 9$

3 0

4 years ago

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Alik [6]

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4 0

3 years ago

I WILL MARK BRAINLIST <br> Which problem can be solved by performing this multiplication? 3/4×8/9

Lana71 [14]

The answer is 4/9

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3 0

3 years ago

Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

finlep [7]

Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

b) $a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

7 0

3 years ago

Solve this equation 2(x + 2)/3 = 6

Naddik [55]

Answer:

Step-by-step explanation:

2(x + 2)/3 = 6

multiply by : 3

2(x + 2) = 18

2x+4 = 18

2x = 18-4=14

x=14/2

x=7

3 0

3 years ago

If sin∅+cos∅ = 1 , find sin∅.cos∅.​

If sin∅+cos∅ = 1 , find sin∅.cos∅.