How would I solve sinø= 1 + cosø !? Please help?

Question

Solve 0
sinø= 1 + cosø

Answer 1

sin(t) = 1 + cos(t)

To solve this, let's move the cos(t) to the left hand side.

sin(t) - cos(t) = 1

Square both sides, to obtain

sin^2(t) - 2sin(t)cos(t) + cos^2(t) = 1

sin^2(t) + cos^2(t) - 2sin(t)cos(t) = 1

Note the identity sine squared x plus cos squared x = 1.

1 - 2sin(t)cos(t) = 1
-2sin(t)cos(t) = 0
2sin(t)cos(t) = 0

Note that 2sin(t)cos(t) = sin(2t) by the double angle identity, so

sin(2t) = 0, and it follows that (note: we have to double our solutions because we have 2t)

2t = {0, pi, 2pi, 3pi}. Dividing both sides (and therefore each term) by 2, we get

t = {0, pi/2, pi, 3pi/2}

Because squaring an equation may add unnecessary solutions, we have to TEST each value.

sin(t) = 1 + cos(t).

Test t = 0: sin(0) = 0, 1 + cos(0) = 1 + 1 = 2.
Reject t = 0.

Test t = pi/2: sin(pi/2) = 1, 1 + cos(pi/2) = 1 + 0 = 1.
This one checks out.

Test t = pi: sin(pi) = 0, 1 + cos(pi) = 1 + (-1) = 0.
This one checks out.

Test t = 3pi/2; sin(3pi/2) = -1, 1 + cos(3pi/2) = 1 + 0 = 1.
Reject t = 3pi/2.

Therefore

t = {pi/2, pi}, or, converted to degrees
t = {90, 180}

Answer 2

sinø= 1 + cosø 0°≤ø≤360°

By the 'auxiliary angle' method

so 1sinø - 1cosø = 1

√[(1² + (-1)²] = √2

So 1/√2 sinø - 1/√2 cosø = 1/√2

Let α be the angle such that sinα = 1/√2 and cosα = 1/√2

So α = 45°

So sinø cosα - cosø sinα = 1/√2

ie sin(ø - α) = 1/√2 0°≤ø≤360°

ie sin(ø - 45°) =1/√2 0° - 45°≤ø - 45°≤360° - 45°

sin(ø - 45°) =1/√2 - 45°≤ø - 45°≤315°

So ø - 45° = 45° 135°

So ø = 90°, 180°

Answer 3

sinφ can never exceed 1. Therefore cosφ must be ≤ 0.
φ = arc-cos0 = 90°, 270°, but sin(270°) = -1, so that solution must be rejected.

There may be other angles for which the equation is true.
sinφ - cosφ = 1
squaring,
sin^2φ - 2 sinφcosφ + cos^2φ = 1
1 - 2 sinφcosφ = 1
2 sinφcosφ = 0
arc-sin(0) = 0°, 180°, 360°, so
φ = 0°, 90°, 180°, 360°

Answer 4

1a. sin? - tan? = 0 sin? - (sin?/cos?) = 0 (sin?/cos?)(cos? - a million) = 0 sin? = 0 => ? = ok*one hundred and eighty for ok being an integer OR cos? - a million = 0 => cos? = a million => ? = ok*360 for ok being an integer with a view to keep the answer between 0 and 360 degrees, we actually favor to consider thoughts for even as ok = 0, a million, 2 (probably). note that plugging in ok > 2 might want to bring about an answer previous 360 for particular. ? = 0, one hundred and eighty, 360 are the really solutions that lie on your period. 1b. 2sin2? = ?2 you want to get ? by itself. sin2? = (?2)/2 2? = 40 5 + ok*360 for ok being an integer OR 2? = 130 5 + ok*360 for ok being an integer So our answer turns into ? = 40 5/2 + ok*one hundred and eighty for ok being an integer OR ? = 130 5/2 + ok*one hundred and eighty for ok being an integer to keep the attitude between 0 and 360 degrees, we merely favor to consider ok = 0, a million ? = 40 5/2, 130 5/2, 405/2, 495/2 are the angles the lie on your period. 1c. sin? = a million + cos? sq. both aspect. (sin?)^2 = (a million + cos?)^2 improve the right aspect. (sin?)^2 = a million + 2cos? + (cos?)^2 bear in mind the in call for identity: (sin?)^2 + (cos?)^2 = a million (sin?)^2 = (sin?)^2 + (cos?)^2 + 2cos? + (cos?)^2 Subtract (sin?)^2 from both aspect. 0 = 2(cos?)^2 + 2cos? Divide 2 from both aspect. 0 = (cos?)^2 + cos? component out cos? from the right aspect. 0 = cos?(cos? + a million) cos? = 0 => ? = ninety + ok*one hundred and eighty the position ok is an integer OR cos? = -a million => ? = one hundred and eighty + ok*360 the position ok is an integer to keep ? contained in the perfect period, we actually extremely favor to consider ok = 0, a million ? = ninety, one hundred and eighty are the really angles that lie on your period.